玲珑学院 1137 Sin your life 【数学】
2017-07-27 20:16
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1137 - Sin your life
Time Limit:1s Memory
Limit:128MByte
Submissions:613Solved:112
DESCRIPTION
给一个正整数nn,求下列表达式的最大值:
(sin(x)+sin(y)+sin(z))[x+y+z=n][x≥1][y≥1][z≥1](sin(x)+sin(y)+sin(z))[x+y+z=n][x≥1][y≥1][z≥1]
为了降低难度,这里的x,y,zx,y,z皆为整数
INPUT
输入只有一行,包含一个正整数n(3≤n≤3∗106)n(3≤n≤3∗106)
OUTPUT
输出一行表示答案,请恰好保留99位小数.(你的答案必须和标准答案完全一样才算通过)
SAMPLE INPUT
3
SAMPLE OUTPUT
2.524412954
sin(x)+sin(y)=2⋅sin(x+y2)⋅cos(x−y2) sin(x)+sin(y)=2·sin(x+y2)·con(x−y2)
sin(x)+sin(y)+sin(z)=2⋅sin(x+y2)⋅cos(x−y2)+sin(n−x−y)
化简后从2-n-1枚举x+y的值。因为x+y的值固定,所以当它的值为偶数时,cos那部分的值最大为1(当x==y时);而当它的值为奇数时,x-y可以等于1,3,5...n-2,
这时加个再判断来取cos那部分的最大值。
积化和差公式:
Time Limit:1s Memory
Limit:128MByte
Submissions:613Solved:112
DESCRIPTION
给一个正整数nn,求下列表达式的最大值:
(sin(x)+sin(y)+sin(z))[x+y+z=n][x≥1][y≥1][z≥1](sin(x)+sin(y)+sin(z))[x+y+z=n][x≥1][y≥1][z≥1]
为了降低难度,这里的x,y,zx,y,z皆为整数
INPUT
输入只有一行,包含一个正整数n(3≤n≤3∗106)n(3≤n≤3∗106)
OUTPUT
输出一行表示答案,请恰好保留99位小数.(你的答案必须和标准答案完全一样才算通过)
SAMPLE INPUT
3
SAMPLE OUTPUT
2.524412954
题目链接:
玲珑学院 1137 Sin your life题目大意:
对于给定的n,求当x、y、z均为整数时sin(x)+sin(y)+sin(z)的最大值。结果保留到小数点后9位。解题思路:
如果枚举x、y的话是太暴力了。我们不妨把要求的式子化简一下,利用和差化积公式进行化简sin(x)+sin(y)+sin(z)=2⋅sin(x+y2)⋅cos(x−y2)+sin(n−x−y)
化简后从2-n-1枚举x+y的值。因为x+y的值固定,所以当它的值为偶数时,cos那部分的值最大为1(当x==y时);而当它的值为奇数时,x-y可以等于1,3,5...n-2,
这时加个再判断来取cos那部分的最大值。
Mycode:
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int MAX = 10005; const int MOD = 1e9+7; const int INF = 0x3f3f3f3f; int n; double ans = -INF, tem, mac = -2; int main() { scanf("%d",&n); for(int i = 2; i < n; ++i) { if(i % 2 == 0) tem = 2.0*sin(i/2.0) + sin(n-i); else { mac = max(mac, cos((i-2.0)/2.0)); tem = 2.0*sin(i/2.0)*mac + sin(n-i); } if(ans < tem) ans = tem; } printf("%.9f\n",ans); return 0; }
和差化积公式:
积化和差公式:
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