Project Euler Problem 53： Combinatoric selections【组合数】

2024-05-17 00:06:48

PE其他解题报告请参考这里,本题答案在留言首条

Combinatoric selections

Problem 53

There are exactly ten ways of selecting three from five, 123451234512345:

123,124,125,134,135,145,234,235,245,and345123, 124, 125, 134, 135, 145, 234, 235, 245, and 345123,124,125,134,135,145,234,235,245,and345

In combinatorics, we use the notation, 5C3=105C_3 = 105C3=10.

In general,
nCr=n!r!n!−r!nC_r = \frac {n!r!}{n!−r!}nCr=n!−r!n!r!,where r ≤ n, n! = n×(n−1)×…×3×2×1, and 0! = 1.
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?

题意：

让你求组合数CnmC_n^mCnm 1 <= n <= 100, 1 <= m <= n，大于一百万的组合数的个数

分析：

根据组合数的性质，暴力跑就行，如果大于一百万，就赋值1000001，这样保证不会爆int

参考代码

#include<bits/stdc++.h>using namespace std;int c[111][111];void init() {c[0][0] = 1;int cnt = 0;for (int i = 1; i <= 100; i++) {for (int j = 0; j <= i; j++) {if (j == 0 || j == i) c[i][j] = 1;else {int t = c[i - 1][j] + c[i - 1][j - 1];if (t > 1000000) c[i][j] = (int)(1e6) + 1, cnt++;else c[i][j] = t;}}}cout << cnt << endl;
}int main(){init();return 0;
}```***阅读好习惯：点赞 + 收藏~***

Project Euler Problem 53： Combinatoric selections【组合数】相关推荐

Project Euler：Problem 53 Combinatoric selections
There are exactly ten ways of selecting three from five, 12345: 123, 124, 125, 134, 135, 145, 234, 2 ...
Project Euler Problem 104 Pandigital Fibonacci ends
Pandigital Fibonacci ends Problem 104 The Fibonacci sequence is defined by the recurrence relation: ...
Project Euler Problem 66
Problem 66 Consider quadratic Diophantine equations of the form: x2 – Dy2 = 1 For example, when D=13 ...
Project Euler Problem 27小结
Project Euler上有很多有意思的问题,刚做到第27题,对这个问题做个小结. Problem 27: Euler有一个著名的方程n^2+n+41,当n=0到39时,方程结果均为质数.如今人们用 ...
Project Euler Problem 27 Quadratic primes
Quadratic primes Problem 27 Euler discovered the remarkable quadratic formula: n2+n+41 It turns out ...
Project Euler Problem 92 Square digit chains
Square digit chains Problem 92 A number chain is created by continuously adding the square of the di ...
Project Euler Problem 25 1000-digit Fibonacci number
1000-digit Fibonacci number Problem 25 The Fibonacci sequence is defined by the recurrence relation: ...
Project Euler Problem 14 Longest Collatz sequence
Longest Collatz sequence Problem 14 The following iterative sequence is defined for the set of posit ...
Project Euler Problem 48: Self powers
Self powers Problem 48 The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits ...

最新文章

热门文章