标签：cst iostream ble freopen div ifd mat abs 链接
题目链接：UVA 10883
题意：给定n个数，每相邻两个数取平均数，重复这个过程直到只剩一个数。求最后的数的值。
思路：可以发现答案应该是\(\frac{\sum_{i=0}^{n-1} C_{n-1}^i * d[i]}{2^{n-1}}\)。难点在于n最大值为50000。
所以求解过程中要用log2计算。
代码如下：
1 #include"cstdio" 2 #include"iostream" 3 #include"cstring" 4 #include"algorithm" 5 #include"cstdlib" 6 #include"vector" 7 #include"set" 8 #include"map" 9 #include"cmath" 10 using namespace std; 11 typedef long long LL; 12 const LL MAXN=50010; 13 const LL MOD=1000000000+7; 14 const LL INF=0x3f3f3f3f; 15 16 // C[n][m]=C[n][m-1]*(n-m+1)/m; 17 double d[MAXN]; 18 int main() 19 { 20 #ifdef LOCAL 21 freopen("in.txt","r",stdin); 22 // freopen("out.txt","w",stdout); 23 #endif 24 int t; 25 scanf("%d",&t); 26 for(int tt=1;tt<=t;tt++) 27 { 28 int n; 29 scanf("%d",&n); 30 double C=0; 31 double ans=0; 32 for(int i=0;i<n;i++) 33 { 34 scanf("%lf",&d[i]); 35 if(i) C+=log2(1.0*(n-i)/i); 36 if(d[i]) ans+=((d[i]>0)?1:-1)*pow(2, ( C + log2(fabs(d[i])) - (n-1) )); 37 } 38 39 printf("Case #%d: %.3lf\n",tt,ans); 40 } 41 return 0; 42 }
标签：cst iostream ble freopen div ifd mat abs 链接
原文地址：http://www.cnblogs.com/zarth/p/6761292.html
