HDU 5063 Operation the Sequence

题目链接

把操作存下来。因为仅仅有50个操作，所以每次把操作逆回去执行一遍，就能求出在原来的数列中的位置。输出就可以

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 100005;
const ll MOD = 1000000007;

int t, n, m, c;
ll a[N];
int op[N], on;
char str[3];

ll pow_mod(ll x, ll k) {
	ll ans = 1;
	while (k) {
		if (k&1) ans = ans * x % MOD;
		x = x * x % MOD;
		k >>= 1;
	}
	return ans;
}

ll solve(int c) {
	ll mul = 1;
	for (int i = on - 1; i >= 0; i--) {
		if (op[i] == 1) {
			if (c > (n + 1) / 2) c = (c - (n + 1) / 2) * 2;
			else c = (c - 1) * 2 + 1;
		} else if (op[i] == 2) c = n - c + 1;
		else mul = mul * 2 % (MOD - 1);
	}
	return pow_mod(a[c], mul);
}

int main() {
	scanf("%d", &t);
	while (t--) {
		on = 0;
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) a[i] = i;
		while (m--) {
			scanf("%s%d", str, &c);
			if (str[0] == ‘O‘) op[on++] = c;
			else printf("%lld\n", solve(c));
		}
	}
	return 0;
}