HDU 6205 card card card && 沈阳网络赛1012 (尺取法)

Problem Description
As a fan of Doudizhu, WYJ likes collecting playing cards very much.
One day, MJF takes a stack of cards and talks to him: let’s play a game and if you win, you can get all these cards. MJF randomly assigns these cards into n heaps, arranges in a row, and sets a value on each heap, which is called “penalty value”.
Before the game starts, WYJ can move the foremost heap to the end any times.
After that, WYJ takes the heap of cards one by one, each time he needs to move all cards of the current heap to his hands and face them up, then he turns over some cards and the number of cards he turned is equal to the penaltyvalue.
If at one moment, the number of cards he holds which are face-up is less than the penaltyvalue, then the game ends. And WYJ can get all the cards in his hands (both face-up and face-down).
Your task is to help WYJ maximize the number of cards he can get in the end.So he needs to decide how many heaps that he should move to the end before the game starts. Can you help him find the answer?
MJF also guarantees that the sum of all “penalty value” is exactly equal to the number of all cards.

Input
There are about 10 test cases ending up with EOF.
For each test case:
the first line is an integer n (1≤n≤106), denoting n heaps of cards;
next line contains n integers, the ith integer ai (0≤ai≤1000) denoting there are ai cards in ith heap;
then the third line also contains n integers, the ith integer bi (1≤bi≤1000) denoting the “penalty value” of ith heap is bi.

Output
For each test case, print only an integer, denoting the number of piles WYJ needs to move before the game starts. If there are multiple solutions, print the smallest one.

Sample Input
5
4 6 2 8 4
1 5 7 9 2

Sample Output
4

题意，有n堆牌，拿第一行数量的牌，也必须要拿出第二行数量的牌，如果数量不够不能拿第一行的牌，问当手中的牌不够拿第二行的牌时，拿过第一行的牌的数量，不过可以执行一个操作就是把第前面的牌拿到最后

一道非常简单的最大字段和，就是把上下两个的差值存起来，存储为再把从n+1到2*n的也存储起来，因为是个环，利用尺取法就行了，哎，这题都没做出来，真是菜

代码如下：

#include<bits/stdc++.h>

using namespace std;
const int MAX = 2000100;//这里要存储为2倍的数量
int a[MAX],b[MAX],value[MAX];
int main(void){
    int n;
    while(scanf("%d",&n) != EOF){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i+n] = a[i];
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&b[i]);
            b[i+n] = b[i];
            value[i] = a[i] - b[i];
            value[i+n] = a[i+n] - b[i+n];
        }
        int res = 0,tem = 0,sumA = 0,sum = 0;
        for(int s=1,t=1;t <= 2*n;t++){
            sumA += a[t];
            sum += value[t];
            if(sumA > tem){//跟新最大值
                tem = sumA;
                res = s-1;
            }
            if(sum < 0){
                sum = sumA = 0;
                s = t+1;
            }
            if(t - s + 1 == n){
                break;
            }
        }
        printf("%d\n",res);
    }

    return 0;
}