http://acm.hdu.edu.cn/showproblem.php?pid=3280
用了简单的枚举。
Equal Sum Partitions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 453 Accepted Submission(s): 337
Problem Description
An equal sum partition of a sequence of numbers is a grouping of the numbers (in the same order as the original sequence) in such a way that each group has the same sum. For example, the sequence: 2 5 1 3 3 7 may be grouped as: (2 5) (1 3 3) (7) to yield an equal sum of 7.
Note: The partition that puts all the numbers in a single group is an equal sum partition with the sum equal to the sum of all the numbers in the sequence.
For this problem, you will write a program that takes as input a sequence of positive integers and returns the smallest sum for an equal sum partition of the sequence.
Note: The partition that puts all the numbers in a single group is an equal sum partition with the sum equal to the sum of all the numbers in the sequence.
For this problem, you will write a program that takes as input a sequence of positive integers and returns the smallest sum for an equal sum partition of the sequence.
Input
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by a decimal integer M, (1 ≤ M ≤ 10000), giving the total number of integers in the sequence. The remaining line(s) in the dataset consist of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.
Output
For each data set, generate one line of output with the following values: The data set number as a decimal integer, a space, and the smallest sum for an equal sum partition of the sequence.
Sample Input
3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1
1 2 1 1 2 1 1 2 1 1
Sample Output
1 7
2 21
3 2
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[];
int main()
{
int i,j,t,n,m,sum,cursum,flag ,ans;
scanf("%d",&t);
while(t--)
{
flag=;
memset(a,,sizeof(a));
scanf("%d%d",&n,&m);
for(i=;i<m;i++)
scanf("%d",&a[i]);
for(i=;i<m;i++)
{
sum=;
for(j=;j<=i;j++)
sum+=a[j];
cursum=;
while(j<m)
{
cursum+=a[j];
if(cursum>sum)
break;
else if(cursum==sum)
{
j++;
if(j==m)
{
printf("%d %d\n",n,sum);
flag=;
}
cursum=;
}
else
j++;
if(flag)
break;
} if(flag)
break;
}
if(i==m)
printf("%d %d\n",n,sum);
}
return ;
}
/*
3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1
*/
区间dp
#include<iostream>
#include<cstdio>
using namespace std;
int dp[][],ans[];
int main()
{
int t,n,m,i,j,k,g,a[];
cin>>t;
while(t--)
{
cin>>n>>m;
ans[]=;
for(i=;i<=m;i++)
{
cin>>a[i];
ans[i]=ans[i-]+a[i];
}
for(k=;k<m;k++)//k不能从1-m，虽然同样个数相同，但是j=2开始，就会使区间减少了一层，
{ //比如i=1，j=2就没有这个区间。
for(i=;i<=m-k;i++)
{
j=i+k;
dp[i][j]=ans[j]-ans[i-];//初始化dp，求出每个区间的和。
for(g=i;g<j;g++)
{//三者的顺序可以随便调换。
if((ans[g]-ans[i-])==dp[g+][j])
dp[i][j]=min(dp[i][j],dp[g+][j]);
if(dp[i][g]==ans[j]-ans[g])
dp[i][j]=min(dp[i][j],dp[i][g]);
if(dp[i][g]==dp[g+][j])
dp[i][j]=min(dp[i][j],dp[i][g]); } }
}
printf("%d %d\n",n,dp[][m]);
} }
/*
3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1
*/