acm之动态规划题目1

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

代码：

#include<stdio.h>
#include<iostream>
using namespace std;
int max(int a,int b){
    return a>b?a:b;
}
int main(){
    int a,b,t=0;
    int c[100000];
    int f[100000];
    int p[100000];
    cin>>a;
    while(a--){
        t++;
        cin>>b;
        for(int i=0;i<b;i++){
            cin>>c[i];
            p[i]=i;
        }
        f[0]=c[0];
        for(int i=1;i<b;i++){
            f[i]=max(c[i],f[i-1]+c[i]);
            if(f[i]==c[i])p[i]=i;
            if(f[i]==c[i]+f[i-1])p[i]=p[i-1];
        }
        int Max=f[0];
        int s=p[0];
        int e=0;
        for(int i=1;i<b;i++){
            if(Max<f[i]){
                Max=f[i];
                s=p[i];
                e=i;
            }   
        }
        cout<<"Case "<<t<<":"<<endl<<Max<<" "<<s+1<<" "<<e+1<<endl;
        if(a)cout<<endl;
    }
    return 0;
}