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.

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

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

Sample Output

1 7

2 21

3 2

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[10005];
int main()
{int i,j,t,n,m,sum,cursum,flag ,ans;scanf("%d",&t);while(t--){flag=0;memset(a,0,sizeof(a));scanf("%d%d",&n,&m);for(i=0;i<m;i++)scanf("%d",&a[i]);for(i=0;i<m;i++){sum=0;for(j=0;j<=i;j++)sum+=a[j];cursum=0;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=1;}cursum=0;}elsej++;if(flag)break;}if(flag)break;}if(i==m)printf("%d %d\n",n,sum);}return 0;
}
/*
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[10005][10005],ans[10005];
int main()
{int t,n,m,i,j,k,g,a[10005];cin>>t;while(t--){cin>>n>>m;ans[0]=0;for(i=1;i<=m;i++){cin>>a[i];ans[i]=ans[i-1]+a[i];}for(k=0;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=1;i<=m-k;i++){j=i+k;dp[i][j]=ans[j]-ans[i-1];//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-1])==dp[g+1][j])dp[i][j]=min(dp[i][j],dp[g+1][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+1][j])dp[i][j]=min(dp[i][j],dp[i][g]);}}}printf("%d %d\n",n,dp[1][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
*/

转载于:https://www.cnblogs.com/cancangood/p/3859488.html

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