題目鏈接：http://www.ifrog.cc/acm/problem/1137
題解：設m=n-z
sin(x)+sin(y)=sin(m-y)+sin(y)利用公式得最大值為sqrt(sin(m)*sin(m)+(1-cos(m))*(1-cos(m)))
所以只要便利一下z就可以了。
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define Pi acos(-1)
using namespace std;
int main() {
int n;
scanf("%d" , &n);
double ans = 0 , Max = -3;
for(int i = 1 ; i <= n ; i++) {
int z = i;
int m = n - z;
Max = max(Max , sin(1.0 * z) + 1.0 * sqrt(sin(1.0 * m) * sin(1.0 * m) + (1.0 - cos(1.0 * m)) * (1.0 - cos(1.0 * m))));
}
printf("%.9lf\n" , Max);
return 0;
}