HDU-6052 To my boyfriend 思维
2017-07-29 15:47
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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 621 Accepted Submission(s): 284
Problem Description
Dear Liao
I never forget the moment I met with you. You carefully asked me: "I have a very difficult problem. Can you teach me?". I replied with a smile, "of course". You replied:"Given a matrix, I randomly choose a sub-matrix, what is the expectation of the number of
**different numbers** it contains?"
Sincerely yours,
Guo
Input
The first line of input contains an integer T(T≤8) indicating the number of test cases.
Each case contains two integers, n and m (1≤n, m≤100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains g_i,j (0≤ g_i,j < n*m).
Output
Each case outputs a number that holds 9 decimal places.
Sample Input
1
2 3
1 2 1
2 1 2
Sample Output
1.666666667
Hint
6(size = 1) + 14(size = 2) + 4(size = 3) + 4(size = 4) + 2(size = 6) = 30 / 18 = 6(size = 1) + 7(size = 2) + 2(size = 3) + 2(size = 4) + 1(size = 6)
Source
2017 Multi-University Training Contest - Team 2
Recommend
liuyiding | We have carefully selected several similar problems for you: 6055 6054 6053 6052 6051
题意:对于一个n*m的方格,每个格子中都包含一种颜色,求出任意一个矩形包含不同颜色的期望。
To my boyfriend
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 621 Accepted Submission(s): 284
Problem Description
Dear Liao
I never forget the moment I met with you. You carefully asked me: "I have a very difficult problem. Can you teach me?". I replied with a smile, "of course". You replied:"Given a matrix, I randomly choose a sub-matrix, what is the expectation of the number of
**different numbers** it contains?"
Sincerely yours,
Guo
Input
The first line of input contains an integer T(T≤8) indicating the number of test cases.
Each case contains two integers, n and m (1≤n, m≤100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains g_i,j (0≤ g_i,j < n*m).
Output
Each case outputs a number that holds 9 decimal places.
Sample Input
1
2 3
1 2 1
2 1 2
Sample Output
1.666666667
Hint
6(size = 1) + 14(size = 2) + 4(size = 3) + 4(size = 4) + 2(size = 6) = 30 / 18 = 6(size = 1) + 7(size = 2) + 2(size = 3) + 2(size = 4) + 1(size = 6)
Source
2017 Multi-University Training Contest - Team 2
Recommend
liuyiding | We have carefully selected several similar problems for you: 6055 6054 6053 6052 6051
题意:对于一个n*m的方格,每个格子中都包含一种颜色,求出任意一个矩形包含不同颜色的期望。
这个题得考虑每一个有颜色的点对整个答案有什么贡献。为了避免重复计算这一种颜色的点,(因为一种颜色可能会有多个方格, 在计算某个方格的贡献时,与它相同颜色且被计算过的方格是不能考虑的),这时可以把这些方格当做障碍物。所以可以指定一个 方向算点对答案的贡献。规定一个计算的顺序,比如按照从上到下,从左到右的顺序给每个同色的点排个序,计算这个颜色的贡 献时,对于第i个点,可以计算包括点i以及包括之后所有同色点的矩形数目,但是不能计算包括i之前的点的矩形数目,这样就可 以不重复的计算。具体处理就是处理上边界,左边界,右边界。然后计算方案数就完了。
//china no.1 #pragma comment(linker, "/STACK:1024000000,1024000000") #include <vector> #include <iostream> #include <string> #include <map> #include <stack> #include <cstring> #include <queue> #include <list> #include <stdio.h> #include <set> #include <algorithm> #include <cstdlib> #include <cmath> #include <iomanip> #include <cctype> #include <sstream> #include <functional> #include <stdlib.h> #include <time.h> #include <bitset> using namespace std; #define pi acos(-1) #define endl '\n' #define srand() srand(time(0)); #define me(x,y) memset(x,y,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); cin.tie(0); #define FOR(x,n,i) for(int i=x;i<=n;i++) #define FOr(x,n,i) for(int i=x;i<n;i++) #define W while #define sgn(x) ((x) < 0 ? -1 : (x) > 0) #define bug printf("***********\n"); typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; const int dx[]={-1,0,1,0,1,-1,-1,1}; const int dy[]={0,1,0,-1,-1,1,-1,1}; const int maxn=1e2+10; const int maxx=1e4+100; const double EPS=1e-7; const int MOD=10000007; #define mod(x) ((x)%MOD); template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} inline int Scan() { int Res=0,ch,Flag=0; if((ch=getchar())=='-')Flag=1; else if(ch>='0' && ch<='9')Res=ch-'0'; while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0'; return Flag ? -Res : Res; } //freopen( "in.txt" , "r" , stdin ); //freopen( "data.out" , "w" , stdout ); //cerr << "run time is " << clock() << endl; int n,m,a[maxn][maxn]; //(下边界方案*左边界方案*右边界方案,上边界已经确定是xi) double solve(int x,int y) { double res=0; int c=a[x][y],L=1,R=m; for(int i=x;i>=1;i--)//上边界 { if(i<x&&a[i][y]==c) break; // 同一列 int l=y,r=y; for(int j=y-1;j>=max(1,L);j--)//左边界 { if(a[i][j]==c) break; l=j; } L=max(l,L); if(i==x) { res+=(LL)(n-x+1LL)*(y-L+1LL)*(R-y+1LL); continue; } for(int j=y+1;j<=min(m,R);j++)//右边界 { if(a[i][j]==c) break; r=j; } R=min(R,r); res+=(LL)(n-x+1LL)*(y-L+1LL)*(R-y+1LL); } return res; } int main() { int t; cin>>t; while(t--) { cin>>n>>m; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j]; double ans=0,mx=0; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { ans+=solve(i,j); mx+=i*j; } printf("%.9lf\n",ans/mx); } //cerr << "run time is " << clock() << endl; }
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