NYOJ 927 The partial sum problem 【DFS】+【剪枝】

2024-05-18 16:16:22

The partial sum problem

时间限制：1000 ms | 内存限制：65535 KB

难度：2

描写叙述

One day,Tom’s girlfriend give him an array A which contains N integers and asked him:Can you choose some integers from the N integers and the sum of them is equal to K.

输入

There are multiple test cases.
Each test case contains three lines.The first line is an integer N(1≤N≤20),represents the array contains N integers. The second line contains N integers,the ith integer represents A[i](-10^8≤A[i]≤10^8).The third line contains an integer K(-10^8≤K≤10^8).

输出

If Tom can choose some integers from the array and their them is K,printf ”Of course,I can!”; other printf ”Sorry,I can’t!”.

例子输入

例子输出

Of course,I can!
Sorry,I can't!

这题非常经典，剪枝的时候要细心。

#include <stdio.h>
#include <stdlib.h>
int n, arr[22], sum, vis[22], ok, count;
const char *sam[] = {"Sorry,I can't!\n", "Of course,I can!\n"};int cmp(const void *a, const void *b){return *(int *)a - *(int *)b;
}void DFS(int k){if(count == sum){ok = 1; return;}for(int i = k; i < n; ++i){if(i && arr[i] == arr[i-1] && !vis[i-1]) //cutcontinue;if(count > sum && arr[i] > 0) return; //cutcount += arr[i]; vis[i] = 1;DFS(i + 1);if(ok) return;count -= arr[i]; vis[i] = 0;}
}int main(){while(scanf("%d", &n) == 1){for(int i = 0; i < n; ++i){scanf("%d", arr + i);vis[i] = 0;}scanf("%d", &sum);qsort(arr, n, sizeof(int), cmp);count = ok = 0; DFS(0);printf(ok ? sam[1] : sam[0]);}return 0;
}

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