HDU 1003 Max Sum (O（N） 算法)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 210088    Accepted Submission(s): 49285

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6大体题意：给你n 个元素的数组，求出一个连续子序列 使得子序列的和最大，并且输出 子序列的起始位置和终止位置！思路：直接遍历数组，thissum += a[i]，当比maxsum 大时，更新maxsum 并且更新左位置和右位置如果thissum < 0肯定不能作为起点，也不是最优的直接把thissum初始化为0，然后更新左位置！
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 100000 + 10;int a[maxn];int l,r,n,k;int solve(){    l = r = k = 1;    int maxsum = -maxn,thissum = 0;    for (int i = 1; i <= n; ++i){        thissum += a[i];         if (thissum > maxsum)maxsum = thissum,r=i,l=k;        if (thissum < 0)thissum = 0,k=i+1;    }    return maxsum;}int main(){    int T,cnt=0;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for (int i = 1; i <= n; ++i){            scanf("%d",&a[i]);        }        int ans = solve();        if (cnt)printf("\n");        printf("Case %d:\n",++cnt);        printf("%d %d %d\n",ans,l,r);    }    return 0;}