This article is an English translation of the NOIP Round 1 (the Qualification Round) problems.↵
[cut]↵
↵
Here are some backgrounds about it.↵
↵
* [user:matthew99,2018-10-201] has written an [Introduction to Chinese OI contests](http://mirror.codeforces.com/blog/entry/51462), but he ignored this for some reason; maybe this exam is too easy for him in his province.↵
* It is literally a "preliminary contest".↵
* Basically, this is a qualification round for NOIP.↵
* This is a written examination.↵
* The exam lasts for 2 hours.↵
* The total score is 100. You may need to get ~80 scores (in provinces like Zhejiang) or ~20 scores (in provinces like Inner Mongolia) to pass.↵
* This translation may be not 100% exact, but you can feel the flavor of the NOIPR1.↵
↵
In part IV and V, the translator modified and formatted the source codes with `clang-format` .↵
↵
### Part I: Single Choice Problems↵
↵
2 score for each problem.↵
↵
1. In the following four numbers in different positional notation, which one is different from the other three?↵
* A. (269)16↵
* B. (617)10↵
* C. (1151)8↵
* D. (1001101011)2↵
↵
2. Which of the following programming language is run by a interpreter?↵
* A. C↵
* B. C++↵
* C. Pascal↵
* D. Python↵
↵
3. When did Chinese Computer Federation founded the NOI competitions? (it was called "National High School Students' Computer Programming Contest")↵
* A. 1983↵
* B. 1984↵
* C. 1985↵
* D. 1986↵
↵
4. Let the depth of a single node to be 0. What is the number of nodes of a full k-ary tree of depth h?↵
* A. $(k^{h+1}-1)/(k-1)$↵
* B. $k^{h-1}$↵
* C. $k^{h}$↵
* D. $(k^{h-1})/(k-1)$↵
↵
5. There is an algorithm whose time-cost function is T(n) = T(n — 1) + n, T(0)=1. What is its time complexity?↵
* A. $O(\log n)$ ↵
* B. $O(n \log n)$ ↵
* C. $O(n)$ ↵
* D. $O(n^2)$↵
↵
6. What is the prefix form (Polish notation) of the expression `a*d-b*c`↵
* A. `a d * b c * -`↵
* B. `- * a d * b c`↵
* C. `a * d - b * c`↵
* D. `- * * a d b c`↵
↵
7. Choose two points A and B randomly on a segment of length 1. What's the expected length of $|AB|$ ?↵
* A. 1 / 2↵
* B. 1 / 3↵
* C. 2 / 3↵
* D. 3 / 5↵
↵
8. Catalan Numbers is $C_n = (2n)! / (n + 1)! / n!$ . Which of the following is wrong?↵
* A. $C_n$ = the number of different binary trees of n+1 nodes↵
* B. $C_n$ = the number of legal bracket sequence of length 2n↵
* C. $C_n$ = the number of possible "pop sequence" if we push 1,2,...,n into a stack↵
* D. $C_n$ = the number of ways to triangulate a convex (n+2)-gon↵
↵
9. Assume there is a machine with a button. Whenever you push that button, it will randomly give you either a red ball or a blue ball with 1/2 probability each. Infinitively many people play with this machine. Each of them push that button until a blue ball is given. Then they will put all their balls into a big box. What is the ratio of the two colors of the balls in the box?↵
* A. 1 : 2↵
* B. 2 : 1↵
* C. 1 : 3↵
* D. 1 : 1↵
↵
10. You need to write a code to compute the number of 1's in a binary form of a number n.↵
Here is the code:↵
↵
```c++↵
int CountBit(int x)↵
{↵
int ret = 0;↵
while (x)↵
{↵
ret++;↵
________;↵
}↵
return ret;↵
}↵
```↵
↵
What should you fill in the blank?↵
* A. `x >>= 1`↵
* B. `x &= x - 1`↵
* C. `x |= x >> 1`↵
* D. `x <<= 1`↵
↵
### Part II: Multiple Choice Problems↵
↵
There are several choices in each problem. You must find them all to get 2 score for each problem in this part.↵
↵
1. What can you take into the exam room in NOIP Round 1?↵
* A. Pen or Pencil↵
* B. Eraser↵
* C. Mobile phone (turned off)↵
* D. Draft paper↵
2. 2-3 trees are a special kind of tree. They satisfy:↵
(1) all non-leaf nodes have 2 or 3 children;↵
(2) all leaf has the same depth.↵
If a 2-3 tree have 10 leaves, how many non-leaf nodes may it have?↵
* A. 5B. 6 C. 7 ↵
* B. 6 ↵
* C. 7 ↵
* D. 8↵
3. Which of the following statements is true?↵
* A. if there is no negative cycle but negative edges in a graph, Dijkstra's algorithm may not calculate the single source shortest paths↵
* B. if there is no negative edge, calling Dijkstra's algorithm multiple times may calculate the shortest paths between any pair of nodes↵
* C. if there is negative cycles, you can call Dijkstra's algorithm one time to calculate single source shortest paths↵
* D. when there is no negative edge, calling Dijkstra's algorithm once cannot be used to calculate the shortest paths between any pair of nodes↵
4. Which of the following property does a tree have?↵
* A. there is no cycle↵
* B. there is only one simple path between any pair of nodes↵
* C. have and only have one simple cycle↵
* D. the number of edges equals to the number of nodes minus one↵
5. Which of the following statements about the Turing Award is true?↵
* A. it is set up by Institute of Electrical and Electronics Engineers (IEEE)↵
* B. the only Chinese who has got the Award is Yao Qizhi↵
* C. its name is from the pioneer of computer science, the British scientist Alan Mathison Turing↵
* D. it's the most famous and most honored award in computer field. It was honored as the "Nobel Prize in computer field".↵
↵
### Part III: Problem Solving↵
↵
5 score for each task.↵
↵
1. There are four friends. let's call them Jia, Yi, Bing and Ding.↵
They are deciding whether they will go out for a hike.↵
(1) if it rains and Yi doesn't go out, then Jia will not go out;↵
(2) if Yi goes out, then Ding will go out;↵
(3) if Bing goes out, Ding must not go out;↵
(4) if Ding doesn't go out and Jia doesn't go out, then Bing won't go out.↵
Now you can see Bing is going out. So:↵
* Does Jia go out? ↵
* Does Yi go out? ↵
* Does Ding go out? ↵
* Is it raining?↵
The last problem is 2 scores; the other 1 score.↵
2. How many pairs of solution does the equation `a*b=(a and b)*(a or b)` have when `a,b` are integers in range [0,31]? (`*` means multiplication, `and, or` are bitwise operations)↵
↵
### Part IV: Read the Source Code and Write the Output↵
↵
5 scores for each task.↵
↵
##### Task 1↵
↵
```c++↵
#include <cstdio>↵
int main() {↵
int x;↵
scanf("%d", &x);↵
int res = 0;↵
for (int i = 0; i < x; ++i) {↵
if (i * i % x == 1) {↵
++res;↵
}↵
}↵
printf("%d", res);↵
return 0;↵
}↵
```↵
↵
The input is: `15`↵
What's the output?↵
↵
##### Task 2↵
↵
```c++↵
#include <cstdio>↵
int n, d[100];↵
bool v[100];↵
int main() {↵
scanf("%d", &n);↵
for (int i = 0; i < n; ++i) {↵
scanf("%d", d + i);↵
v[i] = false;↵
}↵
int cnt = 0;↵
for (int i = 0; i < n; ++i) {↵
if (!v[i]) {↵
for (int j = i; !v[j]; j = d[j]) {↵
v[j] = true;↵
}↵
++cnt;↵
}↵
}↵
printf("%d\n", cnt);↵
return 0;↵
}↵
```↵
↵
The input is: `10 7 1 4 3 2 5 9 8 0 6`↵
What's the output?↵
↵
##### Task 3↵
↵
```c++↵
#include <iostream>↵
using namespace std;↵
string s;↵
long long magic(int l, int r) {↵
long long ans = 0;↵
for (int i = l; i <= r; ++i) {↵
ans = ans * 4 + s[i] - 'a' + 1;↵
}↵
return ans;↵
}↵
int main() {↵
cin >> s;↵
int len = s.length();↵
int ans = 0;↵
for (int l1 = 0; l1 < len; ++l1) {↵
for (int r1 = l1; r1 < len; ++r1) {↵
bool bo = true;↵
for (int l2 = 0; l2 < len; ++l2) {↵
for (int r2 = l2; r2 < len; ++r2) {↵
if (magic(l1, r1) == magic(l2, r2) && (l1 != l2 || r1 != r2)) {↵
bo = false;↵
}↵
}↵
}↵
if (bo) {↵
ans += 1;↵
}↵
}↵
}↵
cout << ans << endl;↵
return 0;↵
}↵
```↵
↵
The input is: `abacaba`↵
What's the output?↵
↵
##### Task 4↵
↵
```c++↵
#include <cstdio>↵
using namespace std;↵
const int N = 110;↵
bool isUse[N];↵
int n, t;↵
int a[N], b[N];↵
bool isSmall() {↵
for (int i = 1; i <= n; ++i)↵
if (a[i] != b[i])↵
return a[i] < b[i];↵
return false;↵
}↵
bool getPermutation(int pos) {↵
if (pos > n) {↵
return isSmall();↵
}↵
for (int i = 1; i <= n; ++i) {↵
if (!isUse[i]) {↵
b[pos] = i;↵
isUse[i] = true;↵
if (getPermutation(pos + 1)) {↵
return true;↵
}↵
isUse[i] = false;↵
}↵
}↵
return false;↵
}↵
void getNext() {↵
for (int i = 1; i <= n; ++i) {↵
isUse[i] = false;↵
}↵
getPermutation(1);↵
for (int i = 1; i <= n; ++i) {↵
a[i] = b[i];↵
}↵
}↵
int main() {↵
scanf("%d%d", &n, &t);↵
for (int i = 1; i <= n; ++i) {↵
scanf("%d", &a[i]);↵
}↵
for (int i = 1; i <= t; ++i) {↵
getNext();↵
}↵
for (int i = 1; i <= n; ++i) {↵
printf("%d", a[i]);↵
if (i == n)↵
putchar('\n');↵
else↵
putchar(' ');↵
}↵
return 0;↵
}↵
```↵
↵
The first input is: `6 10 1 6 4 5 3 2`↵
↵
The second input is: `6 200 1 5 3 4 2 6` ↵
↵
What are the outputs?↵
↵
### Part V: Program Blank Fill-in↵
↵
In this part, you are given several programs with their descriptions, but some expressions in the programs are deleted. Fill in the blanks to make the programs work as described.↵
↵
##### Task 1↵
↵
You have a 1-indexed permutation $P$ of length $n$. We define array $Q$ as: $Q_i$ is the smallest $j$ satisfying $j>i\and P_j>P_i$ . If such $j$ doesn't exist, $Q_i=n+1$ .↵
↵
For example, if $n=5, P=\{1,5,4,2,3\}$, then $Q=\{2,6,6,5,6\}$.↵
↵
The following program reads $P$ and calculates $Q$ with double-linked list. It is guaranteed that $n\leq 10^5$.↵
↵
```c++↵
#include <iostream>↵
using namespace std;↵
const int N = 100010;↵
int n;↵
int L[N], R[N], a[N];↵
int main() {↵
cin >> n;↵
for (int i = 1; i <= n; ++i) {↵
int x;↵
cin >> x;↵
/*fill in(1)*/;↵
}↵
for (int i = 1; i <= n; ++i) {↵
R[i] = /*fill in(2)*/;↵
L[i] = i - 1;↵
}↵
for (int i = 1; i <= n; ++i) {↵
L[/*fill in(3)*/] = L[a[i]];↵
R[L[a[i]]] = R[/*fill in(4)*/];↵
}↵
for (int i = 1; i <= n; ++i) {↵
cout << /*fill in(5)*/ << " ";↵
}↵
cout << endl;↵
return 0;↵
}↵
```↵
↵
##### Task 2↵
↵
Piggy wants to buy $n$ objects. $n\leq 1000$. He can choose from two shops, A and B.↵
↵
Each of these $n$ objects are sold in both shops. If Piggy buy the i-th object at A shop, the cost is a[i]. If Piggy buy the i-th object at B shop, the cost is b[i]. All the prices are no less than 0 and no more than 10000. If the total price in shop A is no less than 50000, everything Piggy buy at A shop will have a 5% discount. That is, if `total_a >= 50000`, `result = total_a * 0.95 + total_b`.↵
↵
Calculate the minimum amount of money does Piggy need to buy the $n$ objects.↵
↵
```c++↵
#include <cstdio>↵
#include <algorithm>↵
using namespace std;↵
const int Inf = 1000000000;↵
const int threshold = 50000;↵
const int maxn = 1000;↵
int n, a[maxn], b[maxn];↵
bool put_a[maxn];↵
int total_a, total_b;↵
double ans;↵
int f[threshold];↵
int main() {↵
scanf("%d", &n);↵
total_a = total_b = 0;↵
for (int i = 0; i < n; ++i) {↵
scanf("%d%d", a + i, b + i);↵
if (a[i] <= b[i])↵
total_a += a[i];↵
else↵
total_b += b[i];↵
}↵
ans = total_a + total_b;↵
total_a = total_b = 0;↵
for (int i = 0; i < n; ++i) {↵
if (/*fill in(1)*/) {↵
put_a[i] = true;↵
total_a += a[i];↵
} else {↵
put_a[i] = false;↵
total_b += b[i];↵
}↵
}↵
if (/*fill in(2)*/) {↵
printf("%.2f", total_a * 0.95 + total_b);↵
return 0;↵
}↵
f[0] = 0;↵
for (int i = 1; i < threshold; ++i)↵
f[i] = Inf;↵
int total_b_prefix = 0;↵
for (int i = 0; i < n; ++i)↵
if (!put_a[i]) {↵
total_b_prefix += b[i];↵
for (int j = threshold - 1; j >= 0; --j) {↵
if (/*fill in(3)*/ >= threshold && f[j] != Inf)↵
ans = min(ans, (total_a + j + a[i]) * 0.95 + /*fill in(4)*/);↵
f[j] = min(f[j] + b[i], j >= a[i] ? /*fill in(5)*/ : Inf);↵
}↵
}↵
printf("%.2f", ans);↵
return 0;↵
}↵
```↵
↵
The answerwill be translated and published maybe tomorrowis being translated.
[cut]↵
↵
Here are some backgrounds about it.↵
↵
* [user:matthew99,2018-10-2
* It is literally a "preliminary contest".↵
* Basically, this is a qualification round for NOIP.↵
* This is a written examination.↵
* The exam lasts for 2 hours.↵
* The total score is 100. You may need to get ~80 scores (in provinces like Zhejiang) or ~20 scores (in provinces like Inner Mongolia) to pass.↵
* This translation may be not 100% exact, but you can feel the flavor of the NOIPR1.↵
↵
In part IV and V, the translator modified and formatted the source codes with `clang-format` .↵
↵
### Part I: Single Choice Problems↵
↵
2 score for each problem.↵
↵
1. In the following four numbers in different positional notation, which one is different from the other three?↵
* A. (269)16↵
* B. (617)10↵
* C. (1151)8↵
* D. (1001101011)2↵
* A. C↵
* B. C++↵
* C. Pascal↵
* D. Python↵
* A. 1983↵
* B. 1984↵
* C. 1985↵
* D. 1986↵
* A. $(k^{h+1}-1)/(k-1)$↵
* B. $k^{h-1}$↵
* C. $k^{h}$↵
* D. $(k^{h-1})/(k-1)$↵
* A. $O(\log n)$ ↵
* B. $O(n \log n)$ ↵
* C. $O(n)$ ↵
* D. $O(n^2)$↵
* A. `a d * b c * -`↵
* B. `- * a d * b c`↵
* C. `a * d - b * c`↵
* D. `- * * a d b c`↵
* A. 1 / 2↵
* B. 1 / 3↵
* C. 2 / 3↵
* D. 3 / 5↵
* A. $C_n$ = the number of different binary trees of n+1 nodes↵
* B. $C_n$ = the number of legal bracket sequence of length 2n↵
* C. $C_n$ = the number of possible "pop sequence" if we push 1,2,...,n into a stack↵
* D. $C_n$ = the number of ways to triangulate a convex (n+2)-gon↵
* A. 1 : 2↵
* B. 2 : 1↵
* C. 1 : 3↵
* D. 1 : 1↵
Here is the code:↵
int CountBit(int x)↵
{↵
int ret = 0;↵
while (x)↵
{↵
ret++;↵
________;↵
}↵
return ret;↵
}↵
```↵
* A. `x >>= 1`↵
* B. `x &= x - 1`↵
* C. `x |= x >> 1`↵
* D. `x <<= 1`↵
↵
### Part II: Multiple Choice Problems↵
↵
There are several choices in each problem. You must find them all to get 2 score for each problem in this part.↵
↵
1. What can you take into the exam room in NOIP Round 1?↵
* A. Pen or Pencil↵
* B. Eraser↵
* C. Mobile phone (turned off)↵
* D. Draft paper↵
2. 2-3 trees are a special kind of tree. They satisfy:↵
(1) all non-leaf nodes have 2 or 3 children;↵
(2) all leaf has the same depth.↵
If a 2-3 tree have 10 leaves, how many non-leaf nodes may it have?↵
* A. 5
* B. 6 ↵
* C. 7 ↵
* D. 8↵
3. Which of the following statements is true?↵
* A. if there is no negative cycle but negative edges in a graph, Dijkstra's algorithm may not calculate the single source shortest paths↵
* B. if there is no negative edge, calling Dijkstra's algorithm multiple times may calculate the shortest paths between any pair of nodes↵
* C. if there is negative cycles, you can call Dijkstra's algorithm one time to calculate single source shortest paths↵
* D. when there is no negative edge, calling Dijkstra's algorithm once cannot be used to calculate the shortest paths between any pair of nodes↵
4. Which of the following property does a tree have?↵
* A. there is no cycle↵
* B. there is only one simple path between any pair of nodes↵
* C. have and only have one simple cycle↵
* D. the number of edges equals to the number of nodes minus one↵
5. Which of the following statements about the Turing Award is true?↵
* A. it is set up by Institute of Electrical and Electronics Engineers (IEEE)↵
* B. the only Chinese who has got the Award is Yao Qizhi↵
* C. its name is from the pioneer of computer science, the British scientist Alan Mathison Turing↵
* D. it's the most famous and most honored award in computer field. It was honored as the "Nobel Prize in computer field".↵
↵
### Part III: Problem Solving↵
↵
5 score for each task.↵
↵
1. There are four friends. let's call them Jia, Yi, Bing and Ding.↵
They are deciding whether they will go out for a hike.↵
(1) if it rains and Yi doesn't go out, then Jia will not go out;↵
(2) if Yi goes out, then Ding will go out;↵
(3) if Bing goes out, Ding must not go out;↵
(4) if Ding doesn't go out and Jia doesn't go out, then Bing won't go out.↵
Now you can see Bing is going out. So:↵
* Does Jia go out? ↵
* Does Yi go out? ↵
* Does Ding go out? ↵
* Is it raining?↵
The last problem is 2 scores; the other 1 score.↵
2. How many pairs of solution does the equation `a*b=(a and b)*(a or b)` have when `a,b` are integers in range [0,31]? (`*` means multiplication, `and, or` are bitwise operations)↵
↵
### Part IV: Read the Source Code and Write the Output↵
↵
5 scores for each task.↵
↵
##### Task 1↵
↵
```c++↵
#include <cstdio>↵
int main() {↵
int x;↵
scanf("%d", &x);↵
int res = 0;↵
for (int i = 0; i < x; ++i) {↵
if (i * i % x == 1) {↵
++res;↵
}↵
}↵
printf("%d", res);↵
return 0;↵
}↵
```↵
↵
The input is: `15`↵
What's the output?↵
↵
##### Task 2↵
↵
```c++↵
#include <cstdio>↵
int n, d[100];↵
bool v[100];↵
int main() {↵
scanf("%d", &n);↵
for (int i = 0; i < n; ++i) {↵
scanf("%d", d + i);↵
v[i] = false;↵
}↵
int cnt = 0;↵
for (int i = 0; i < n; ++i) {↵
if (!v[i]) {↵
for (int j = i; !v[j]; j = d[j]) {↵
v[j] = true;↵
}↵
++cnt;↵
}↵
}↵
printf("%d\n", cnt);↵
return 0;↵
}↵
```↵
↵
The input is: `10 7 1 4 3 2 5 9 8 0 6`↵
What's the output?↵
↵
##### Task 3↵
↵
```c++↵
#include <iostream>↵
using namespace std;↵
string s;↵
long long magic(int l, int r) {↵
long long ans = 0;↵
for (int i = l; i <= r; ++i) {↵
ans = ans * 4 + s[i] - 'a' + 1;↵
}↵
return ans;↵
}↵
int main() {↵
cin >> s;↵
int len = s.length();↵
int ans = 0;↵
for (int l1 = 0; l1 < len; ++l1) {↵
for (int r1 = l1; r1 < len; ++r1) {↵
bool bo = true;↵
for (int l2 = 0; l2 < len; ++l2) {↵
for (int r2 = l2; r2 < len; ++r2) {↵
if (magic(l1, r1) == magic(l2, r2) && (l1 != l2 || r1 != r2)) {↵
bo = false;↵
}↵
}↵
}↵
if (bo) {↵
ans += 1;↵
}↵
}↵
}↵
cout << ans << endl;↵
return 0;↵
}↵
```↵
↵
The input is: `abacaba`↵
What's the output?↵
↵
##### Task 4↵
↵
```c++↵
#include <cstdio>↵
using namespace std;↵
const int N = 110;↵
bool isUse[N];↵
int n, t;↵
int a[N], b[N];↵
bool isSmall() {↵
for (int i = 1; i <= n; ++i)↵
if (a[i] != b[i])↵
return a[i] < b[i];↵
return false;↵
}↵
bool getPermutation(int pos) {↵
if (pos > n) {↵
return isSmall();↵
}↵
for (int i = 1; i <= n; ++i) {↵
if (!isUse[i]) {↵
b[pos] = i;↵
isUse[i] = true;↵
if (getPermutation(pos + 1)) {↵
return true;↵
}↵
isUse[i] = false;↵
}↵
}↵
return false;↵
}↵
void getNext() {↵
for (int i = 1; i <= n; ++i) {↵
isUse[i] = false;↵
}↵
getPermutation(1);↵
for (int i = 1; i <= n; ++i) {↵
a[i] = b[i];↵
}↵
}↵
int main() {↵
scanf("%d%d", &n, &t);↵
for (int i = 1; i <= n; ++i) {↵
scanf("%d", &a[i]);↵
}↵
for (int i = 1; i <= t; ++i) {↵
getNext();↵
}↵
for (int i = 1; i <= n; ++i) {↵
printf("%d", a[i]);↵
if (i == n)↵
putchar('\n');↵
else↵
putchar(' ');↵
}↵
return 0;↵
}↵
```↵
↵
The first input is: `6 10 1 6 4 5 3 2`↵
↵
The second input is: `6 200 1 5 3 4 2 6` ↵
↵
What are the outputs?↵
↵
### Part V: Program Blank Fill-in↵
↵
In this part, you are given several programs with their descriptions, but some expressions in the programs are deleted. Fill in the blanks to make the programs work as described.↵
↵
##### Task 1↵
↵
You have a 1-indexed permutation $P$ of length $n$. We define array $Q$ as: $Q_i$ is the smallest $j$ satisfying $j>i\and P_j>P_i$ . If such $j$ doesn't exist, $Q_i=n+1$ .↵
↵
For example, if $n=5, P=\{1,5,4,2,3\}$, then $Q=\{2,6,6,5,6\}$.↵
↵
The following program reads $P$ and calculates $Q$ with double-linked list. It is guaranteed that $n\leq 10^5$.↵
↵
```c++↵
#include <iostream>↵
using namespace std;↵
const int N = 100010;↵
int n;↵
int L[N], R[N], a[N];↵
int main() {↵
cin >> n;↵
for (int i = 1; i <= n; ++i) {↵
int x;↵
cin >> x;↵
/*fill in(1)*/;↵
}↵
for (int i = 1; i <= n; ++i) {↵
R[i] = /*fill in(2)*/;↵
L[i] = i - 1;↵
}↵
for (int i = 1; i <= n; ++i) {↵
L[/*fill in(3)*/] = L[a[i]];↵
R[L[a[i]]] = R[/*fill in(4)*/];↵
}↵
for (int i = 1; i <= n; ++i) {↵
cout << /*fill in(5)*/ << " ";↵
}↵
cout << endl;↵
return 0;↵
}↵
```↵
↵
##### Task 2↵
↵
Piggy wants to buy $n$ objects. $n\leq 1000$. He can choose from two shops, A and B.↵
↵
Each of these $n$ objects are sold in both shops. If Piggy buy the i-th object at A shop, the cost is a[i]. If Piggy buy the i-th object at B shop, the cost is b[i]. All the prices are no less than 0 and no more than 10000. If the total price in shop A is no less than 50000, everything Piggy buy at A shop will have a 5% discount. That is, if `total_a >= 50000`, `result = total_a * 0.95 + total_b`.↵
↵
Calculate the minimum amount of money does Piggy need to buy the $n$ objects.↵
↵
```c++↵
#include <cstdio>↵
#include <algorithm>↵
using namespace std;↵
const int Inf = 1000000000;↵
const int threshold = 50000;↵
const int maxn = 1000;↵
int n, a[maxn], b[maxn];↵
bool put_a[maxn];↵
int total_a, total_b;↵
double ans;↵
int f[threshold];↵
int main() {↵
scanf("%d", &n);↵
total_a = total_b = 0;↵
for (int i = 0; i < n; ++i) {↵
scanf("%d%d", a + i, b + i);↵
if (a[i] <= b[i])↵
total_a += a[i];↵
else↵
total_b += b[i];↵
}↵
ans = total_a + total_b;↵
total_a = total_b = 0;↵
for (int i = 0; i < n; ++i) {↵
if (/*fill in(1)*/) {↵
put_a[i] = true;↵
total_a += a[i];↵
} else {↵
put_a[i] = false;↵
total_b += b[i];↵
}↵
}↵
if (/*fill in(2)*/) {↵
printf("%.2f", total_a * 0.95 + total_b);↵
return 0;↵
}↵
f[0] = 0;↵
for (int i = 1; i < threshold; ++i)↵
f[i] = Inf;↵
int total_b_prefix = 0;↵
for (int i = 0; i < n; ++i)↵
if (!put_a[i]) {↵
total_b_prefix += b[i];↵
for (int j = threshold - 1; j >= 0; --j) {↵
if (/*fill in(3)*/ >= threshold && f[j] != Inf)↵
ans = min(ans, (total_a + j + a[i]) * 0.95 + /*fill in(4)*/);↵
f[j] = min(f[j] + b[i], j >= a[i] ? /*fill in(5)*/ : Inf);↵
}↵
}↵
printf("%.2f", ans);↵
return 0;↵
}↵
```↵
↵
The answer
