HDU 4923 Room and Moor
先依照类似于 110 1110 11110分区
在每个区间内取平均值
假设后一个区间取得的值大于前一个区间,则将两个区间合并,取平均值
(感谢LUKE队长以及ASCE的思路
#include<cstdio> #include<cmath> #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<map> #define pi acos(-1.0) #define eps 1e-8 #define ll long long #define L 1000050 #define N 100050 #define Mod 1000000007 #define M 99999999999 #define clr(a) memset(a,0,sizeof(a)) #pragma comment(linker, "/STACK:102400000,102400000") using namespace std; template<class T> inline bool read(T &n) { T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } template <class T> inline void write(T n) { if(n < 0) { putchar('-'); n = -n; } int len = 0,data[20]; while(n) { data[len++] = n%10; n /= 10; } if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } double s[N][2]; int main() { int i,j,k,n; int cas=1; int a,b; int t; double x; int num=0; for(read(t); t--;) { read(n); s[0][0]=1; s[0][1]=1; num=0; for(i=1; i<=n; i++) { read(x); s[num][0]=x; s[num][1]=1; num++; while(num>=2) { if((s[num-1][0]/s[num-1][1])>(s[num-2][0]/s[num-2][1]))break; s[num-2][0]+=s[num-1][0]; s[num-2][1]+=s[num-1][1]; num--; } } double res=0; for(i=0; i<num; i++) { double tmp=s[i][0]/s[i][1]; res+=tmp*tmp*(s[i][1]-s[i][0])+(1-tmp)*(1-tmp)*(s[i][0]); } printf("%.6lf\n",res); //printf("Case #%d: %d\n",cas++,res); } return 0; }