题目
Max Sum
题面
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.
Examples
Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Output
Case 1:
14 1 4
Case 2:
7 1 6
思路
这个题目可以把计算过的前部分看成一个整体（一个数字），如果发现这个整体的和小于0，就从下一个数字重新开始加（因为前面的和为负数，我们要求最大和，加上的话反而和会变小），咋这个过程中，不断更新和的最大值，即可得到答案。
代码
#include<stdio.h>
int main()
{
long long max,n,sum;
//freopen("F:/y.txt","r",stdin);
int T;
scanf("%d",&T);
for(int t=1;t<=T;t++)
{
scanf("%lld",&n);
int l1=1,l=1,r=1;
sum=0;
max=0;
int f=1;
long long x;
for(int i=1;i<=n;i++)
{
scanf("%lld",&x);
if(i==1)
max=x;
sum+=x;
if(sum>max)
{
max=sum;
l=l1;
r=i;
}
if(sum<0)
{
sum=0;
l1=i+1;
}
}
printf("Case %d:\n",t);
printf("%lld %d %d\n",max,l,r);
if(t!=T)
printf("\n");
}
return 0;
}
