（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: [email protected]) // 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 = 1; if(r > n / 2) r = n - r; for(i = n; i >= n - r + 1; i--) { s *= i; } for(i = 1; i <= r; i++) { s /= i; } return s; } int main() { int i, j, s; s = 0; for(i = 23; i <= 100; i++) { j = 3; while(combinatoric(i, j) < 1000000) j++; if(i % 2) { s += (i / 2 - j + 1) * 2; //利用组合数的对称性，分奇偶两种情况 } else { s += (i / 2 - j) * 2 + 1; } } printf("%d\n", s); return 0; }
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Answer:
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4075 |