The partial sum problem
时间限制:
1000 ms | 内存限制:
65535 KB
难度:
2
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描述
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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.
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输入
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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!”. 样例输入
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4 1 2 4 7 13 4 1 2 4 7 15
样例输出
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Of course,I can! Sorry,I can't!
来源
- 原创 上传者
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TC_李远航
ac代码
#include<string.h> #include<math.h> #include<stdio.h> #include<stdlib.h> int cmp(const void *a,const void *b) { return *(int *)a-*(int *)b; } int a[30],n,sum,w,v[30]; void dfs(int k,int cot) { int i; if(cot==sum) { w=1; return; } if(w) return; for(i=k;i<n;i++) { if(i&&a[i]==a[i-1]&&!v[i-1]) continue; if(cot>sum&&a[i]>0) return; v[i]=1; dfs(i+1,cot+a[i]); if(w) return; v[i]=0; } } int main() { while(scanf("%d",&n)!=EOF) { int i; for(i=0;i<n;i++) scanf("%d",&a[i]); scanf("%d",&sum); qsort(a,n,sizeof(a[0]),cmp); w=0; memset(v,0,sizeof(v)); dfs(0,0); if(w) printf("Of course,I can!\n"); else printf("Sorry,I can't!\n"); } }
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There are multiple test cases.