Max Sum
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
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1050;
const int INF = 0x3f3f3f3f;
#define PI 3.1415927
int a;
int t;
int n, m;
struct Node
{
int x, y, sum;
};
Node ans, tt;//ans存的是答案,tt存的是临时数据
int main()
{
scanf("%d", &t);
getchar();//读掉多余的换行
int a=1;
while(t--)
{
ans = {1, 1, -INF};//初始化两个结构体,-INF是为了防止整个数列全部是负数
tt = {1,0,0};
scanf("%d", &n);
for(int i = 1; i <= n; ++i)
{
int temp;
scanf("%d", &temp);
tt.sum += temp;//录入一个新的变量,临时数据的sum加上新变量的值
++tt.y;//录入一个新的变量,右边界+1
if(tt.sum > ans.sum)//临时变量里的sum>答案里的sum,更新答案
{
ans = tt;
}
if(tt.sum < 0)//当临时变量里的sum<0,易知该数列+后面任何一个数列(有可能只有一个数)都只会使后面哪个数列的值变小,
{ //所以可以直接舍去前面的数列,改变左边界继续计算后面的数列
tt.sum = 0;
tt.x = i+1;
tt.y = i;
}
}
printf("Case %d:\n",a++);
printf("%d %d %d\n", ans.sum, ans.x, ans.y);
if(t!=0)
printf("\n");
}
return 0;
}