1003 Max Sum（水题）

Max Sum

Problem 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

虽然是道水题…..但刚看题还以为用线段树做….想了想线段树…..后来发现什么鬼啊，直接从头遍历一遍，如果遇到这个时候的总和小于0就将此节点更新为起点，如果这个时候的总和大于遍历过的值的最大和，就将max更新为sum,然后把这个点设为结束点，注意的是刚开始不能把max设为0，因为更新的最大值可能为负值
AC代码：

#include<bits/stdc++.h>using namespace std;const int INF=0x3f3f3f3f;int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int T,temp,n;    cin>>T;    for(int k=1;k<=T;k++) {        cin>>n;        int max=-INF,sum=0;        int start=1,end=1,start_t=1;        for(int i=1;i<=n;i++) {            cin>>temp;            sum+=temp;            if(sum>max) {                max=sum;                end=i;                start=start_t;            }            if(sum<0) {                sum=0;                start_t=i+1;            }        }        if(k!=1) cout<<endl;        cout<<"Case "<<k<<":"<<endl;        cout<<max<<" "<<start<<" "<<end<<endl;    }    return 0;}

四级考完….该好好准备期末和刷题了