题意：有n堆牌，ai表示每堆牌的牌数，bi表示每堆牌的e数量的牌翻过来使其面朝下，然后继续操作下一堆牌。问操作开始前，最少重复进行多少次挪的操作可以使手中最终拿的牌的数量最多。
分析：从第一堆开始，边拿边统计，直到游戏停止，然后从停止的下一张牌开始重新边拿边统计，尺取法思想。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const LL MOD = 998244353;
const double pi = acos(-1.0);
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN], b[MAXN];
bool vis[MAXN];
int main(){
int n;
while(scanf("%d", &n) == 1){
memset(vis, false, sizeof vis);
for(int i = 1; i <= n; ++i){
scanf("%d", &a[i]);
}
for(int i = 1; i <= n; ++i){
scanf("%d", &b[i]);
}
int st = 1, ma = 0, ans = 0;
while(!vis[n]){
int sum = 0;
int cnt = 0;
int tmp;
for(int i = 0; i < n; ++i){
tmp = st + i;
if(tmp > n) tmp -= n;
vis[tmp] = true;
cnt += a[tmp];
sum += a[tmp];
if(sum >= b[tmp]){
sum -= b[tmp];
}
else break;
}
if(cnt > ma){
ma = cnt;
ans = st - 1;
}
st = tmp + 1;
}
printf("%d\n", ans);
}
return 0;
}