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GDUFE ACM-1020
The Rascal Triangle
Time Limit: 1000ms
Problem Description:
The Rascal Triangle definition is similar to that of the Pascal Triangle. The rows are numbered from the top starting with 0. Each row n contains n+1 numbers indexed from 0 to n. Using R(n,m) to indicte the index m item in the index n row: R(n,m) = 0 for n < 0 OR m < 0 OR m > n The first and last numbers in each row(which are the same in the top row) are 1: R(n,0) = R(n,n) = 1 The interior values are determined by (UpLeftEntry*UpRightEntry+1)/UpEntry(see the parallelogram in the array below): R(n+1, m+1) = (R(n,m) * R(n,m+1) + 1)/R(n-1,m)
Write a program which computes R(n,m) the
element of the
row of the Rascal Triangle.
Input:
The first line of input contains a single integer P, (1 <= P <= 1000), which is the number of data sets that follow. Each data set is a single line of input consisting of 3 space separated decimal integers. The first integer is data set number, N. The second integer is row number n, and the third integer is the index m within the row of the entry for which you are to find R(n,m) the Rascal Triangle entry (0 <= m <= n <= 50,000).
Output:
For each data set there is onr line of output. It contains the data set number, N, followed by a single space which is then followed by thr Rascal Triangle entry R(n,m) accurate to the integer value.
Sample Input:
5 1 4 0 2 4 2 3 45678 12345 4 12345 9876 5 34567 11398
Sample Output:
1 1 2 5 3 411495886 4 24383845 5 264080263
Source:
2011 Greater New York Regional
这道题就是找规律:
写成这样就可以发现规律了
AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 long long n,m,k,b,j,c,i,t,a[50005]; 7 scanf("%lld",&t); 8 while(t--) 9 { 10 scanf("%lld%lld%lld",&k,&n,&m); 11 if(m==0||m==n) 12 printf("%lld 1\n",k); 13 else 14 { 15 b=n-3; 16 a[0]=1; 17 a[1]=n; 18 for(i=2;b>0;i++) 19 { 20 a[i]=a[i-1]+b; 21 b=b-2; 22 } 23 if(n%2==1) 24 { 25 if(m<=n/2) 26 printf("%lld %lld\n",k,a[m]); 27 else 28 { 29 j=(n+1)/2; 30 c=(m+1)%j; 31 m=j-c;//后半部分,比如n=7时,a[4]=a[3],a[5]=a[2],a[6]=a[1]; 32 printf("%lld %lld\n",k,a[m]); 33 } 34 } 35 else 36 { 37 if(m<=n/2) 38 printf("%lld %lld\n",k,a[m]); 39 else 40 { 41 j=n/2; 42 c=m%j; 43 m=j-c;//类似上面 44 printf("%lld %lld\n",k,a[m]); 45 } 46 } 47 } 48 } 49 return 0; 50 }
另附上高人的代码(我没有想到这个思路):
1 #include <stdio.h> 2 //规律:(n-m)*m+1 3 int main(){ 4 int t,N,n,m; 5 scanf("%d",&t); 6 while(t--){ 7 scanf("%d %d %d",&N,&n,&m); 8 printf("%d %d\n",N,1+(n-m)*m); 9 } 10 return 0; 11 }
GDUFE ACM-1020
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