Codeforces 555D Case of a Top Secret

Case of a Top Secret

感觉除了两个点在那循环的部分， 其他时候绳子的长度每次变为一半一下， 就变成了Log(l)步。。

然后就暴力找就好啦， 循环的部分取个模。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, m, a[N], x[N], id[N];

bool cmp(const int& a, const int& b) {
    return x[a] < x[b];
}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        x[i] = a[i];
        id[i] = i;
    }
    sort(id + 1, id + 1 + n, cmp);
    sort(x + 1, x + 1 + n);
    while(m--) {
        int p, l;
        scanf("%d%d", &p, &l);
        while(1) {
            int be = p, len = 0;
            int it1 = upper_bound(x + 1, x + 1 + n, a[p] + l) - x - 1;
            len += abs(x[it1] - a[p]);
            l -= abs(x[it1] - a[p]);
            p = id[it1];
            int it2 = lower_bound(x + 1, x + 1 + n, a[p] - l) - x;
            len += abs(x[it2] - a[p]);
            l -= abs(x[it2] - a[p]);
            p = id[it2];
            int ed = p;
            if(be == ed && be == id[it1]) {
                break;
            }
            if(be == ed) {
                l %= len;
            }
        }
        printf("%d
", p);
    }
    return 0;
}

/*
*/