uva10759 - Dice Throwing(掷骰子)
2013-05-19 23:19
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题意很简单,,,
三步求答案———(1)求分母,(2)求分子,求分子,(3)分母的最大公约数,
其中最难求的就是分子,,,
此处用的是动规,,,,思路挺简单的,记忆化搜索,
代码如下:
#include <cstdio>
#include <cmath>
#include <cstring>
int n, x;
long long dp[24+5][150+5];
long long m, z, g;
long long gcd(long long a, long long b)
{
return b==0?a:gcd(b,a%b);
}
long long my_pow(int a, int k)
{
long long ans = 1;
for(int i = 0; i < k; i++) ans*=a;
return ans;
}
void ddp()
{
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= x; j++)
for(int k = 1; k <= 6; k++) if(j>=k)
dp[i][j] += dp[i-1][j-k];
z = m;
for(int i = 1; i < x; i++)
z-=dp
[i];
}
int main ()
{
while(scanf("%d%d",&n,&x),n+x)
{
m = my_pow(6,n);
ddp();
if(z==0) {printf("0\n"); continue; }
if(z==m) {printf("1\n"); continue; }
g = gcd(m,z);
printf("%lld/%lld\n",z/g,m/g);
}
return 0;
}
三步求答案———(1)求分母,(2)求分子,求分子,(3)分母的最大公约数,
其中最难求的就是分子,,,
此处用的是动规,,,,思路挺简单的,记忆化搜索,
代码如下:
#include <cstdio>
#include <cmath>
#include <cstring>
int n, x;
long long dp[24+5][150+5];
long long m, z, g;
long long gcd(long long a, long long b)
{
return b==0?a:gcd(b,a%b);
}
long long my_pow(int a, int k)
{
long long ans = 1;
for(int i = 0; i < k; i++) ans*=a;
return ans;
}
void ddp()
{
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= x; j++)
for(int k = 1; k <= 6; k++) if(j>=k)
dp[i][j] += dp[i-1][j-k];
z = m;
for(int i = 1; i < x; i++)
z-=dp
[i];
}
int main ()
{
while(scanf("%d%d",&n,&x),n+x)
{
m = my_pow(6,n);
ddp();
if(z==0) {printf("0\n"); continue; }
if(z==m) {printf("1\n"); continue; }
g = gcd(m,z);
printf("%lld/%lld\n",z/g,m/g);
}
return 0;
}
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