标签：result more str bsp enc sequence tput 14. return
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Code
#include<stdio.h>
int main()
{
int t,n,i,j;
int a;
int sum,maxsum,start,end,temp;
scanf("%d",&t);
i=1;
while(i<=t)
{
scanf("%d",&n);
sum=0,maxsum=-1001,start=0,end=0,temp=1;
for(j=0;j<n;j++)
{
scanf("%d",&a);
sum += a;
if(sum>maxsum)
{
maxsum=sum;
start=temp;
end=j+1;
}
if(sum<0)
{
sum=0;
temp=j+2;
}
}
printf("Case %d:\n",i);
printf("%d %d %d\n",maxsum,start,end);
if(i!=t)
printf("\n");
i++;
}
return 0;
}
Max Sum
标签：result more str bsp enc sequence tput 14. return
原文地址：https://www.cnblogs.com/joannasblog/p/9361658.html
