Project Euler:Problem 76 Counting summations
It is possible to write five as a sum in exactly six different ways:
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
How many different ways can one hundred be written as a sum of at least two positive integers?
#include <iostream>
using namespace std; int c = 0;//累划分数
void p(int n, int a[], int m)//m表示每一种划分的加数的个数
{
int i;
if (n == 0)
{
c++;
//int i;
//for (i = 0; i < m - 1; i++)
// cout << a[i] << "+";
//cout << a[m - 1] << endl;
}
else
for (i = n; i >= 1; i--)
{
if (m == 0 || i <= a[m - 1])//要保证下一个划分因子不大于上一个划分因子
{
a[m] = i;
p(n - i, a, m + 1);
}
}
} void main(void)
{
int n;
int a[200] = { 0 };//存储整数n的划分
printf("输入要被划分的整数: ");
cin >> n;
p(n, a, 0);
cout << "整数" << n << "的划分数是:" << c-1 << "种。" << endl;
system("pause");
}
Project Euler:Problem 76 Counting summations的更多相关文章
- Project Euler:Problem 77 Prime summations
It is possible to write ten as the sum of primes in exactly five different ways: 7 + 3 5 + 5 5 + 3 + ...
- Project Euler:Problem 55 Lychrel numbers
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. Not all numbers produce palindr ...
- Project Euler:Problem 63 Powerful digit counts
The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is ...
- Project Euler:Problem 86 Cuboid route
A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the o ...
- Project Euler:Problem 87 Prime power triples
The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is ...
- Project Euler:Problem 89 Roman numerals
For a number written in Roman numerals to be considered valid there are basic rules which must be fo ...
- Project Euler:Problem 93 Arithmetic expressions
By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four ari ...
- Project Euler:Problem 39 Integer right triangles
If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exact ...
- Project Euler:Problem 28 Number spiral diagonals
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is forme ...
随机推荐
- android 传递 类对象 序列化 Serializable
public class Song implements Serializable { /** * */ private static final long serialVersionUID = 64 ...
- BZOJ2754 SCOI2012喵星球上的点名
绝世好题. 正当我犹豫不决时,hzwer说:“MAP!!!” 没错这题大大的暴力,生猛的stl,贼基尔爽,,ԾㅂԾ,, 由于我们求点名在名字中的子串个数,所以将点名建AC自动机,记录节点属于哪次点名, ...
- [BZOJ2429][HAOI2006]聪明的猴子(最小生成树)
性质:最小生成树上任意两点间的最大边权,一定是这两点间所有路径的最大边权中最小的.证明显然. #include<cstdio> #include<cstring> #inclu ...
- HDU 1057 What Are You Talking About trie树 简单
http://acm.hdu.edu.cn/showproblem.php?pid=1075 题意 : 给一个单词表然后给一些单词,要求翻译单词表中有的单词,没有则直接输出原单词. 翻译文段部分get ...
- SpringBoot返回结果如果为null或空值不显示处理方法
第一种方法:自定义消息转换器 @Configuration public class WebMvcConfig extends WebMvcConfigurerAdapter{ // /** // * ...
- poj 3624 Charm Bracelet 背包DP
Charm Bracelet Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3624 Descripti ...
- pm2-web
A web based monitor for PM2. Multiple hosts With the release of 0.11 pm2 no longer uses TCP sockets ...
- .NET:C#的匿名委托 和 Java的匿名局部内部类
背景 这几天重温Java,发现Java在嵌套类型这里提供的特性比较多,结合自身对C#中匿名委托的理解,我大胆的做了一个假设:Java的字节码只支持静态嵌套类,内部类.局部内部类和匿名局部内部类都是编译 ...
- mysql TO_DAYS()函数
TO_DAYS(date)给定一个日期date, 返回一个天数 (从年份0开始的天数 ). 例: select TO_DAYS(NOW()); +----------------+ | TO_DA ...
- android:Layout_weight的深刻理解
最近写Demo,突然发现了Layout_weight这个属性,发现网上有很多关于这个属性的有意思的讨论,可是找了好多资料都没有找到一个能够说的清楚的,于是自己结合网上资料研究了一下,终于迎刃而解,写出 ...