给你两个数t,n
接下来输入n个数字
让你输出所有数字相加等于n的组合
4 6 4 3 2 2 1 1
t n
4
3+1
2+2
2+1+1
Sample Input
4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25 0 0
Sample Output
Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25
其实就是个简单的dfs,一开始我还以为什么背包问题,结果发现想太多;
#include <iostream> using namespace std; int n,m; int a[15],lu[15]; int cnt = 0; void dfs(int k,int t); bool flag; int main() { int i,j; while(scanf("%d%d",&m,&n) && m) { flag = false; for(i=0;i<n;++i) scanf("%d",a+i); printf("Sums of %d: ",m); dfs(-1,m); if(!flag) printf("NONE "); } } void dfs(int k,int t) { if(t == 0) { flag = true; printf("%d",lu[0]); for(int i=1; i<cnt; ++i) { printf("+%d",lu[i]); } printf(" "); return ; } if(t < 0) return ; for(int j=k+1; j<n; ++j) { if(t >= a[j]) { lu[cnt++] = a[j]; dfs(j,t-a[j]); while(j < n-1 && a[j] == a[j+1]) //防止重复的元素进入 j ++; cnt --; } } }