1137

http://www.ifrog.cc/acm/problem/1137

和差化积公式，

变成2 * sin((x + y) / 2) * cos((x - y) / 2) + sin(n - (x + y))

然后很重要的一个就是cos(x) = cos(-x)

这样，枚举x + y的取值i，当x + y取值是i的时候，x - y的取值也是有固定的规律，具体就是

i是偶数的时候，0、2、...、这样吧

i是奇数的时候，1、3......、这样吧

然后就要分情况取值，

当sin((x + y) / 2)小于0，就要取一个min去和它乘，否则娶个大的，

数组开大了TLE，不是很懂。

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
double dp[2][2];
void work() {
    int n;
    scanf("%d", &n);
    dp[0][0] = dp[0][1] = 1;
    dp[1][0] = dp[1][1] = cos(0.5);
    double ans = -1.0;
    for (int i = 2; i <= n - 1; ++i) {
        double res = sin(i * 0.5);
        if (res < 0) {
            if (i & 1) {
                ans = max(ans, 2 * res * dp[1][0] + sin(n - i));
            } else ans = max(ans, 2 * res * dp[0][0] + sin(n - i));
        } else {
            if (i & 1) {
                ans = max(ans, 2 * res * dp[1][1] + sin(n - i));
            } else ans = max(ans, 2 * res * dp[0][1] + sin(n - i));
        }
        res = cos(i);
        dp[0][0] = min(dp[0][0], res);
        dp[0][1] = max(dp[0][1], res);
        dp[1][0] = min(dp[1][0], res);
        dp[1][1] = max(dp[1][1], res);
    }
    printf("%0.9f
", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 3e6 + 20;
double dp[maxn][2];
void work() {
    int n;
    scanf("%d", &n);
    dp[0][0] = dp[0][1] = cos(0);
    dp[1][0] = dp[1][1] = cos(0.5);
//    for (int i = 2; i <= n; ++i) {
//        dp[i][0] = min(dp[i - 2][0], cos(i * 0.5));
//        dp[i][1] = max(dp[i - 2][1], cos(i * 0.5));
//    }
    double ans = -1.0;
    for (int i = 2; i <= n - 1; ++i) {
        double res = sin(i * 0.5);
        if (i & 1) {
            ans = max(ans, 2 * res * dp[i - 2][0] + sin(n - i));
        } else {
            ans = max(ans, 2 * res * dp[i - 2][1] + sin(n - i));
        }
        res = cos(i * 0.5);
        dp[i][0] = min(dp[i - 2][0], res);
        dp[i][1] = max(dp[i - 2][1], res);
    }
    printf("%0.9f
", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}