HDU1003 Max Sum(DP)

Max SumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 228524    Accepted Submission(s): 53767Problem 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 6

12月12日唉,那时候电脑也是在这一天买的2333.不能对不起自己的电脑，要让它发光发热。求最大子段和，注意初始化(-1000~1000)，所以初始化的值不是0,要在-1000以下，设置为-1001就可以，注意一下格式。

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define MOD 1000000007#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define Pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define inf 0x3f3f3f3f#define Pi 4.0*atan(1.0)#define Sqrt(x) (x)*(x)#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long ll;typedef unsigned long long ull;const double eps = 1e-12;const int maxn = 1000000+10;using namespace std;inline int read(){    int x(0),f(1);    char ch=getchar();    while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int dp[maxn];int main(){    fin;    int t,n;    t=read();    for(int cnt=1;cnt<=t;++cnt){        int Max=-1001,cur=0;        n=read();        int st,ed,tmp;        st=ed=tmp=1;        for(int i=1;i<=n;++i){            dp[i]=read();            cur+=dp[i];            if(cur>Max){                Max=cur;                st=tmp;                ed=i;            }            if(cur<0){                cur=0;                tmp=i+1;            }        }        printf("Case %d:\n%d %d %d\n",cnt,Max,st,ed);        if(cnt!=t){            cout<<endl;        }    }    return 0;}