HDU 5063 Operation the Sequence（暴力）

HDU 5063 Operation the Sequence

题目链接

把操作存下来，由于只有50个操作，所以每次把操作逆回去运行一遍，就能求出在原来的数列中的位置，输出即可

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;}