Light oj 1064 Throwing Dice (概率dp)

题意：给定n个骰子和一个x，要求出用这些骰子投出大于等于x的概率。要求最简。

分析：先预处理算出i个骰子得到的和为j的种数。然后用gcd函数约分。

dp[i+1][j+k]=sigma(dp[i][j]) (1<=k<=6)

AC代码1：

#include<cstdio>#include<iostream>#include<cstring>typedef unsigned long long ull;using namespace std;__int64 dp[30][155];__int64 gcd(__int64 x,__int64 y){    return y?gcd(y,x%y):x;}int main(){    __int64 up,down,g;    int T,n,x,i,j,k;    for(i=1;i<=6;i++)        dp[1][i]=1;    for(i=1;i<25;i++)    {        for(j=i;j<=6*i;j++)        {            for(k=1;k<=6;k++)                dp[i+1][j+k]+=dp[i][j];        }    }    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        scanf("%d%d",&n,&x);        if(x>n*6)        {            printf("Case %d: 0\n",cas);            continue;        }        if(x<=n)        {            printf("Case %d: 1\n",cas);            continue;        }        up=0,down=1;        for(i=x;i<=n*6;i++)            up+=dp[n][i];        for(j=0;j<n;j++) down*=6;        g=gcd(up,down);        printf("Case %d: %lld/%lld\n",cas,up/g,down/g);    }    return 0;}

AC代码2：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;__int64 dp[25][155];__int64 gcd(__int64 x,__int64 y){    return y?gcd(y,x%y):x;}int main(){    __int64 up,down,g;    int T,n,x;    for (int i = 1; i <= 24; i ++)for (int j = 1; j <= 150; j ++) {if (i == 1 && j <= 6)dp[i][j] = 1;for (int k = 1; k <= 6; k ++) {if (j >= k)dp[i][j] += dp[i - 1][j - k];}}    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        scanf("%d%d",&n,&x);        up=0,down=0;        for(int i=n;i<=n*6;i++)        {            down+=dp[n][i];            if(x<=i)                up+=dp[n][i];        }        if(up==down)            printf("Case %d: 1\n",cas);        else if(up==0)            printf("Case %d: 0\n",cas);        else        {            g=gcd(up,down);            printf("Case %d: %lld/%lld\n",cas,up/g,down/g);        }    }    return 0;}