hdu6205

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

54 6 2 8 41 5 7 9 2

 

Sample Output

4
Hint
[pre]For the sample input:+ If WYJ doesn't move the cards pile, when the game starts the state of cards is:4 6 2 8 41 5 7 9 2WYJ can take the first three piles of cards, and during the process, the number of face-up cards is 4-1+6-5+2-7. Then he can't pay the the "penalty value" of the third pile, the game ends. WYJ will get 12 cards.+ If WYJ move the first four piles of cards to the end, when the game starts the state of cards is:4 4 6 2 82 1 5 7 9WYJ can take all the five piles of cards, and during the process, the number of face-up cards is 4-2+4-1+6-5+2-7+8-9. Then he takes all cards, the game ends. WYJ will get 24 cards.It can be improved that the answer is 4.**huge input, please use fastIO.**[/pre]
题意：n堆扑克牌，从第1堆开始，当到某一堆使朝上的牌少于朝下的牌时，游戏结束，否则，取走所有的牌，在游戏开始之前可以移动若干堆，把最前面的牌移到最后面，问能取走的最多的牌数时，需要移动几堆
思路：直接复制一遍，求一遍前缀和，暴力即可。
代码：
#include<bits/stdc++.h>using namespace std;const int maxn = 1e6 + 5;int sum[maxn * 2], a[maxn * 2], b[maxn * 2];struct FastIO{    static const int S = 1310720;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) exit(0);        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for(; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(int x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;int main(){    while(1)    {        int n;        n = io.xuint();        //scanf("%d",&n);        int flag = 0;        for(int i = 0; i < maxn; i++) sum[i] = 0;        for(int i = 1; i <= n; i++) a[i]=io.xuint();        for(int i = 1; i <= n; i++) b[i]=io.xuint();        for(int i = n + 1; i <= 2 * n; i++) a[i] = a[i - n];        for(int i = n + 1; i <= 2 * n; i++) b[i] = b[i - n];        for(int i = 1; i <= 2 * n; i++) sum[i] += (sum[i - 1] + a[i] - b[i]);        for(int i = 1; i <= n; i++)        {            for(int j = i; j <= i + n; j++)            {                if(sum[j] - sum[i - 1] < 0)                {                    i = j;                    break;                }                if(j == n)                {                    //printf("%d\n",i-1);                    io.wint(i - 1);                    io.wchar('\n');                    flag = 1;                }            }            if(flag) break;        }    }    return 0;}