hdu-1003 Max Sum （最大子段和）

http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 202859    Accepted Submission(s): 47418

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

题意：求一个数列的最大子序列和

简单的动态规划，用dp[i]来记录i为终点的序列的最大子段和

如果dp[i-1]>0  dp[i] = dp[i-1] + a[i]

否则dp[i] = a[i]

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
int a[110000], dp[110000];
int main(){
    int T, n, i, j, x, y, cas, xx, yy;
    cin>>T;
    for(cas=1; cas<=T; cas++){
        cin>>n;
        for(i=1; i<=n; i++)
            cin>>a[i];
        int mmax = -100000;
        memset(dp, 0, sizeof(dp));
        dp[1] = a[1];
        x = 1, y = 1;
        xx = 1, yy = 1;
        for(i=1; i<=n; i++){
            if(dp[i-1]>=0){
                dp[i] = dp[i-1] + a[i];
                yy = i;
            }
            else{
                dp[i] = a[i];
                xx = i;
                yy = i;
            }
            if(mmax <= dp[i]){
                mmax = dp[i];
                y = yy;
                x = xx;
            }
        }
//        for(i=1; i<=n; i++)
//            cout<<dp[i]<<" ";
//        cout<<endl;
        printf("Case %d:\n",cas);
        printf("%d %d %d\n",mmax,x,y);
        if(cas!=T)
            cout<<endl;
    }
    return 0;
}

 

还可以穷举

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int a[110000];
int main(){
    int T, n, i, j, sum, k, ans, cas, x, y;
    cin>>T;
    for(cas=1; cas<=T; cas++){
        cin>>n;
        for(i=1; i<=n; i++)
            cin>>a[i];
        sum = 0;
        ans = -1000000;
        k = 1;
        for(i=1; i<=n; i++){
            sum += a[i];
            if(sum>ans){
                ans = sum;
                x = k;
                y = i;
            }
            if(sum<0){
                sum = 0;
                k = i+1;
            }
        }
        printf("Case %d:\n",cas);
        printf("%d %d %d\n",ans, x, y);
        //cout<<cas<<" "<<T<<endl;
        if(cas!=T)
            printf("\n");
    }
    return 0;
}

这里的sum就相当于dp[i]