Combinations |
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN:
Compute the EXACT value of:
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long.
For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input and Output
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.
The output from this program should be in the form:
N things taken M at a time is C exactly.
Sample Input
100 6 20 5 18 6 0 0
Sample Output
100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.
#include"stdio.h" double fun1(int a) { int i; double s=1; i=a; while(i) { s*=i; i--; } return s; } double fun2(int c,int d) { int j; double s=1; for(j=d+1;j<=c;j++) s*=j; return s; } int main() { int m,n; double c,s1,s2; while(1) {scanf("%d %d",&n,&m); if(n==0&&m==0) break; s1=fun1(n-m);//调用函数计算n-m的累乘; s2=fun2(n,m);//调用函数计算m到n的累乘; c=s2/s1; printf("%d things taken %d at a time is %.0lf exactly.\n",n,m,c); } return 0; }