HDU 1003Max Sum 动态规划 经典题 最大子序列和

Max SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 203541    Accepted Submission(s): 47572Problem DescriptionGiven 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.InputThe 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).OutputFor 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 Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5Sample OutputCase 1:14 1 4Case 2:7 1 6AuthorIgnatius.L

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

#include <iostream>#include <cstdio>#include <climits>using namespace std;int A[100001],MAXA[100001];void Handle(int N)///状态转移方程 MAXA(i) = max{F[i],F[i]+MAXA(i-1)}{    int Rb=0,Begin=0,Re=0,End=0,MAX=INT_MIN;MAXA[0]=-9999;///丫的，不能设置成INT_MIN ，会在第一次就因为溢出变成正数///设置变量全部从0开始防止出出现全0    for(int i=1;i<=N;i++)    {        if(A[i]>A[i]+MAXA[i-1]) {Begin=End=i; MAXA[i]=A[i];}        else                    {MAXA[i]=A[i]+MAXA[i-1];End++;}        if(MAX < MAXA[i])       {Rb=Begin;Re = End;MAX = MAXA[i];}    }    printf("%d %d %d\n",MAX,Rb,Re);}int main(){    //freopen("F:\\test.txt","r",stdin);    int T;scanf("%d",&T);    for(int i=1,N;i<=T&&scanf("%d",&N);i++)    {        if(i>=2) putchar('\n');        printf("Case %d:\n",i);        for(int j=1;j<=N;j++) scanf("%d",&A[j]);Handle(N);    }}

来源： http://acm.hdu.edu.cn/viewcode.php?rid=17251534