思维

把数组扩大一倍，然后找连续的大于0的最长的一段即可，注意结尾特判一下。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 2000005;int a[N], b[N], c[N], n;int main(){    while(~scanf("%d", &n)){        for(int i = 1; i <= n; i ++) a[i] = read();        for(int i = 1; i <= n; i ++) b[i] = read();        for(int i = 1; i <= n; i ++) c[i] = a[i] - b[i];        for(int i = n + 1; i <= 2 * n; i ++) c[i] = a[i - n] - b[i - n];        int sum = 0, index = -1, maxi = -INF, ans = -1, len = 0;        for(int i = 1; i <= 2 * n; i ++){            if(index == -1 && c[i] < 0) continue;            else if(index == -1) sum += c[i], index = i, len ++;            else if(sum + c[i] >= 0){                sum += c[i], len ++;                if(i == 2 * n){                    if(len > maxi) maxi = len, ans = index;                }            }            else if(sum + c[i] < 0){                if(len > maxi) maxi = len, ans = index;                sum = 0, len = 0, index = -1;            }            if(maxi == n) break;        }        printf("%d\n", ans - 1);    }    return 0;}