标签：code 大于 mon void lightoj type test bic names
n common cubic dice are thrown. What is the probability that the sum of all thrown dice is at least x?
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each test case contains two integers n (1 ≤ n < 25) and x (0 ≤ x < 150). The meanings of n and x are given in the problem statement.
Output
For each case, output the case number and the probability in ‘p/q‘ form where p and q are relatively prime. If q equals 1 then print p only.
Sample Input
7
3 9
1 7
24 24
15 76
24 143
23 81
7 38
Sample Output
Case 1: 20/27
Case 2: 0
Case 3: 1
Case 4: 11703055/78364164096
Case 5: 25/4738381338321616896
Case 6: 1/2
Case 7: 55/4665
题解：dp[ i ][ j ]表示 i 个骰子掷出和为 j 的方案数。
所以，dp[ i ][ j ]=dp[ i ][ j ]+dp[ i - 1 ][ j - k ](1<=k<j)
1 #define INF 1e8 2 #include<cstdio> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 typedef long long ll; 8 9 int n,x; 10 ll sum; 11 ll dp[25][150]; 12 13 ll gcd(ll a,ll b){ 14 return b==0?a:gcd(b,a%b); 15 } 16 17 void solve(){ 18 sum=1; 19 memset(dp,0,sizeof(dp)); 20 dp[0][0]=1; 21 for(int i=1;i<=n;i++){ 22 sum=sum*6; 23 for(int j=1;j<=i*6;j++) 24 for(int k=1;k<=6;k++) 25 if(j>=k) dp[i][j]=dp[i][j]+dp[i-1][j-k]; //因为dp[0][0]=1,所以j一定要大于等于k，否则不能更新dp[i][j]的值 26 } 27 } 28 29 int main() 30 { int kase; 31 cin>>kase; 32 for(int t=1;t<=kase;t++){ 33 cin>>n>>x; 34 solve(); 35 ll ans=0; 36 for(int i=x;i<=n*6;i++) ans+=dp[n][i]; 37 ll temp=gcd(sum,ans); 38 if(ans==0) printf("Case %d: 0\n",t); 39 else if(ans==sum) printf("Case %d: 1\n",t); 40 else printf("Case %d: %lld/%lld\n",t,ans/temp,sum/temp); 41 } 42 }
标签：code 大于 mon void lightoj type test bic names
原文地址：http://www.cnblogs.com/zgglj-com/p/7341045.html
