The partial sum problem

時間限制：1000?ms ?|? 內存限制：65535?KB 難度：2 描寫敘述

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.? 輸入

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).

輸出

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!”.

例子輸入

4 1 2 4 7 13 4 1 2 4 7 15

例子輸出

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

這題非常經典，剪枝的時候要細心。

#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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