01月05, 2019

hdu1003

题目描述:

Given a sequence a,a,a......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.

输入：

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).

输出：

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.

样本输入：

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

Case 1:

14 1 4

Case 2:

7 1 6

代码：

#include <stdio.h>

int main()
{
int n,m,t;
int sum,max;
int x,y,X,Y;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&m);
sum=-1001;
max=-1001;
for(int j=1;j<=m;j++){
scanf("%d",&t);
if(sum+t<t){
sum=t;
x=j;
y=j;
}
else{
sum+=t;
++y;
}
if(max<sum){
max=sum;
X=x;
Y=y;
}
}
printf("Case %d:\n",i);
printf("%d %d %d\n",max,X,Y);
if(i!=n)
printf("\n");
}
return 0;
}

-- EOF --