codeforces 863D
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原题链接
我的思路:
一开始看到操作,还以为是线段树,吓得我根本就准备溜了。后来想想,能不能直接模拟,算一算发现好像不太能这么做,但是把当前的x还原成在
#include <iostream>#include <cstdio>const int maxn = 2e5 + 20;int n, q, m;int a[maxn], t[maxn], l[maxn], r[maxn];using namespace std;int main() { scanf("%d%d%d", &n, &q, &m); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } for(int i = 1; i <= q; i++) { scanf("%d%d%d", &t[i], &l[i], &r[i]); } while(m--) { int x; scanf("%d", &x); for(int i = q; i > 0; i--) { if(l[i] <= x && x <= r[i]) { if(t[i] == 2) { x = l[i] + r[i] - x; } else { if(x > l[i]) { x--; } else { x = r[i]; } } } } printf("%d ", a[x]); }}
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