解题思路:
利用动态规划方法求解最大子列和,对应输入数据a[i],会有数据dp[i]。数组dp中的每个元素dp[i],表示以a[i]结尾的最大连续子列和。遍历dp数组,可以找出子列和最大值。
if (dp[i-1] >=0){
dp[i] = dp[i-1] + a[i];
}
else{
dp[i] = a[i]
}
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
int r[100001];
int main()
{
//freopen("in.txt","r",stdin);
int cs;
r[0] = -1000000;
scanf("%d",&cs);
int n;
for(int t=1;t<=cs;t++){
scanf("%d",&n);
int s=0,e=0,m=-1000000;
int a,b,tmp;
for(int i=1;i<=n;i++){
scanf("%d",&r[i]);
if(r[i-1]>=0){
r[i]+=r[i-1];
b=i;
tmp = r[i];
}
else{
a = b = i;
tmp = r[i];
}
if(tmp>m){
m = tmp;
s = a;
e = b;
}
}
if(t!=1){
printf("
");
}
printf("Case %d:
%d %d %d
",t,m,s,e);
}
//system("pause");
return 0;
}