HDU 5570(balls-期望)

有n(<=1000)个球，共有m(<=1000)种颜色，第i个球的颜色为j的概率为 ai,jai,1+ai,2+...+ai,m
​对于第 i 种颜色，若有 x 个球，对答案的贡献为 x2 ​​。
问答案的期望
已知 ai,j

转自Bc：
考虑贡献 x2 ​​即为颜色相同的对数，因此只要考虑任意两个球颜色相同的概率，将这些概率加起来就是答案了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iomanip> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) 
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int n,m;
#define MAXN (1000+10)
int a[MAXN][MAXN];
int su[MAXN];
double p[MAXN][MAXN];
int main()
{
// freopen("hdu5570.in","r",stdin);
// freopen(".out","w",stdout);

    while (cin>>n>>m) {
        MEM(su)
        For(i,n) For(j,m) a[i][j]=read(),su[i]+=a[i][j];
        For(i,n) For(j,m) p[i][j]=(double)a[i][j] / su[i];

        double ans=0;


        For(j,m) {
            double sump=0;
            For(i,n) sump+=p[i][j];
            For(i,n) ans+=p[i][j]+(sump-p[i][j])*p[i][j];   

        }
        printf("%.2lf\n",ans);
    }
    return 0;
}