HDU 1003(水DP)

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#include<stdio.h>#include<iostream>#include<math.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<deque>#include<vector>#include<list>#include<map>#include<queue>#include<set>#include<stack>typedef long long LL;typedef unsigned long long uLL;using namespace std;//算法分析：求最大字段和，d[i]表示以i结尾（字段和中包含i）在a[1..i]上的最大和，//d[i]=(d[i-1]+a[i]>a[i])?d[i-1]+a[i]:a[i];max={d[i],1<=i<=n};int a[100010];int main() {    int T,N,k=0,position1,bin,en,mx,sum; cin>>T;    while (T--) {        k++; cin>>N;        for (int i=0; i<N; ++i) {            scanf ("%d",&a[i]);        }        position1=bin=en=0;        mx=sum=a[0];        for (int i=1; i<N; ++i) {            if (sum+a[i]<a[i]) {                sum=a[i]; position1=i;                //若是nowsum+a[i]<a[i],表示当前读入的数据比之前存储的加上当前的还大                //说明可以在当前另外开始记录,更新nowsum=a[i],位置记录改到i            } else {sum+=a[i];}            if (sum>mx) {mx=sum; bin=position1; en=i;}        }        printf ("Case %d:\n%d %d %d\n",k,mx,bin+1,en+1);        if (T) { puts (""); }    }    return 0;}
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