hdoj--1258--Sum It Up(dfs)
Sum It Up
Total Submission(s): 5534 Accepted Submission(s): 2890
and 2+1+1.(A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.
t will be a positive integer less than 1000, n will be an integer between 1 and 12(inclusive), and x1,...,xn will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order,
and there may be repetitions.
A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums
with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distince; the same sum connot appear twice.
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
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
题意:从m个数中取若干个数,是的这若干个数的和为n,输出每一个方案
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[100],n,m,a[100];
int ans;
void dfs(int pre,int x,int r)
{
if(pre==n)
{
ans++;
printf("%d",a[0]);
for(int i=1;i<r;i++)
printf("+%d",a[i]);
printf("\n");
}
else
{
for(int i=x;i<m;i++)
{
if(pre+num[i]<=n)
{
a[r]=num[i];
dfs(pre+num[i],i+1,r+1);
while(i+1<m&&num[i]==num[i+1]) //去重
i++;
}
}
}
}
int cmp(int a,int b)
{
return a>b;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(m==0&&n==0) break;
memset(num,0,sizeof(num));
memset(a,0,sizeof(a));
for(int i=0;i<m;i++)
scanf("%d",&num[i]);
sort(num,num+m,cmp);
printf("Sums of %d:\n",n);
ans=0;
dfs(0,0,0);
if(ans==0)
printf("NONE\n");
}
return 0;
}
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