Bzoj2134:单选错位

题面

Bzoj

Sol

\(i\)道题选对的概率就是\(\frac{min(a[i-1], a[i])}{a[i]*a[i-1]}\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 5);
const int Zsy(1e8 + 1);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, a[_];
double ans;

int main(RG int argc, RG char* argv[]){
	n = Input(); RG int A = Input(), B = Input(), C = Input(); a[1] = Input();
	for(RG int i = 2; i <= n; ++i) a[i] = (1LL * a[i - 1] * A % Zsy + B) % Zsy;
	for(RG int i = 1; i <= n; ++i) a[i] = a[i] % C + 1;
	a[0] = a[n];
	for(RG int i = 1; i <= n; ++i) ans += 1.0 * min(a[i - 1], a[i]) / (1.0 * a[i] * a[i - 1]);
	printf("%.3lf\n", ans);
	return 0;
}
posted @ 2018-02-13 20:35  Cyhlnj  阅读(98)  评论(0编辑  收藏  举报