Time Limit:1000MS
Memory Limit:32768KB
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.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
以下是代码：
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
int main(){
int n,tn;
int news,s,e,t;
int maxn,sum;
cin>> n;
for(int i=1;i<=n;i++){
sum = 0;
maxn = -20000;
news = s = e = 1;
scanf("%d",&tn);
for(int j=1;j<=tn;j++){
scanf("%d",&t);
sum+= t ;
if(sum > maxn){ //目前的和大于记录和，更新和及起点终点
maxn = sum;
e = j;
s = news;
}
if(sum < 0){//目前的和小于0，则将起点定为下一个数，和为0
sum=0;
news=j+1;
}
}
printf("Case %d:\n%d %d %d\n",i,maxn,s,e);
if(i!=n)printf("\n");
}
} |