(Problem 53)Combinatoric selections
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, 5C3 = 10.
In general,

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
How many, not necessarily distinct, values of nCr, for 1 n
100, are greater than one-million?
题目大意:
从五个数12345中选出三个数一共有十种方法:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
在组合数学中我们用5C3 = 10来表示.
n = 23时产生第一个超过一百万的数: 23C10 = 1144066.
对于nCr, 1 n
100,有多少超过100万的值?包括重复的在内。
- //(Problem 53)Combinatoric selections
- // Completed on Fri, 14 Feb 2014, 07:20
- // Language: C11
- //
- // 版权所有(C)acutus (mail: acutus@126.com)
- // 博客地址:http://www.cnblogs.com/acutus/
- #include<stdio.h>
- #include<math.h>
- long long combinatoric(int n, int r) //计算组合数的函数
- {
- int i;
- long long s = ;
- if(r > n / ) r = n - r;
- for(i = n; i >= n - r + ; i--) {
- s *= i;
- }
- for(i = ; i <= r; i++) {
- s /= i;
- }
- return s;
- }
- int main()
- {
- int i, j, s;
- s = ;
- for(i = ; i <= ; i++) {
- j = ;
- while(combinatoric(i, j) < ) j++;
- if(i % ) {
- s += (i / - j + ) * ; //利用组合数的对称性,分奇偶两种情况
- } else {
- s += (i / - j) * + ;
- }
- }
- printf("%d\n", s);
- return ;
- }
Answer:
|
4075 |
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