NYOJ 927 The partial sum problem 【DFS】+【剪枝】
阿新 • • 發佈:2017-06-16
har int 輸出 amp lines cmp dsm sim pri
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]) //cut continue; if(count > sum && arr[i] > 0) return; //cut count += 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; }
NYOJ 927 The partial sum problem 【DFS】+【剪枝】