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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 892 Accepted Submission(s):
590
1 #include<cstring> 2 #include<cstdio> 3 using namespace std; 4 #define LL long long 5 LL num,n,len; 6 char str[15]; 7 void DFS(LL x, LL sum) 8 {//x是取的字符串的长度, sum是当前的值 9 if(x == len)//当取得字符串长度==字符串总串长度时 10 { 11 if(sum==n)//当前结果== n时 12 num++; 13 return ; 14 } 15 LL k=0;//核心的东东 短短的几句 却是指数级的搜索量啊 16 for(int i=x;i<len;i++)//这里刚开始写成了从0开始,哎合适能搜到头啊 17 { 18 k=k*10+str[i]-‘0‘; 19 20 //加减两种可能 21 DFS(i+1, sum+k); 22 if(x != 0)//第一个数字前不能加“-”号 23 DFS(i+1, sum-k); 24 } 25 } 26 int main() 27 { 28 while(scanf("%s %lld",str,&n)!=EOF) 29 { 30 len=strlen(str); 31 num = 0; 32 DFS(0,0); 33 printf("%lld\n", num); 34 } 35 return 0; 36 }
杭电2266 How Many Equations Can You Find【DFS】
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原文地址:http://www.cnblogs.com/123tang/p/5934123.html