Problem A: Equal Sum Partitions
Equal Sum PartitionsTime Limit: 1 Sec Memory Limit:
128 MB
Submit: 36 Solved: 27
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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
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
Sample Output
1 7 2 21 3 2
dp题目真费脑子啊。
#include<stdio.h>
#include<string.h>
int s[1007],dp[1007][1007];
int main()
{
int t,i,j,k;
scanf("%d",&t);
for(int textcase=1; textcase<=t; textcase++)
{
memset(dp,0,sizeof(dp));
int n,m,sum,flag,flag1=0,max;
scanf("%d%d",&n,&m);
for(i=1; i<=m; i++)
{
scanf("%d",&s[i]);
}
for(j=1; j<=m; j++)
{
dp[1][j]=s[j]+dp[1][j-1];
max=dp[1][j];
flag=j;
//printf("\n");
for(k=flag+1; k<=m; k++)
{
dp[flag+1][k]=s[k]+dp[flag+1][k-1];
//printf("%d\n",dp[flag+1][k]);
if(k==m)
{
if(dp[flag+1][m]!=max)
break;
else
{
flag1=1;
break;
}
}
if(dp[flag+1][k]==max)
{
flag=k;
}
}
if(flag1==1)
break;
}
printf("%d %d\n",textcase,max);
}
return 0;
}