The partial sum problem (DFS)nyoj
The partial sum problem
時間限制:1000 ms | 記憶體限制:65535 KB
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!”.
樣例輸入
4 1 2 4 7 13 4 1 2 4 7 15
樣例輸出
Of course,I can! Sorry,I can't!
描述
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.
程式碼:
#include <iostream> #include <cstdio> #include <cstring> using namespace std;
int sum,a[30],k,ok,n;//a陣列用來儲存每個數字,ok存放是否存在題目要求的組合
void dfs(int x)//深搜 { if(sum>k)//當sum比k大的情況,就不用再往深處搜尋了 return ; if(sum==k)//當sum==k,也就是加起來的和存在和k相等的時候,把ok賦值為1; { ok=1; return ; } for (int i = x; i <= n; i++){ sum += a[i];//加上某一個數 dfs(i + 1);//遞迴 sum -= a[i];//把加上的這個數減去 } }
int main() { while(~scanf("%d",&n)){ for(int i=1; i<=n; i++){ scanf("%d",&a[i]); } scanf("%d",&k); ok=0,sum=0; dfs(1); if(ok) cout<<"Of course,I can!"<<endl; else cout<<"Sorry,I can't!"<<endl; } return 0; }