算法题中的常见数学逻辑问题C++

一、 最大公约数的写法:

int gcd(int a, int b){
    return !b ? a : gcd(b, a % b);
}

二、 最小公倍数的写法:

//实质是在得到最大公约数的基础上求解
得到最大公约数d后,最小公倍数等于ab/d

int lcm(int a, int b){
    return a / gcd(a, b) * b;
}

三、分数的四则运算:

//1.分数的表示与化简
struct Fraction{
    int up, down;
};
Fraction reduction(Fraction result){
    if(result.down < 0){
        result.up = -result.up;
        result.down = -result.down;
    }
    if(result.up == 0){
        result.down = 1;
    }else{
        int d = gcd(abs(result.up), abs(result.down));
        result.up /= d;
        result.down /= d;
    }
    return result;
}

//2.分数加法
Fraction add(Fraction f1, Fraction f2){
    Fraction result;
    result.up = f1.up * f2.down + f1.down * f2.up;
    result.down = f1.down * f2.down;
    return reduction(result);
}

//3.分数减法
Fraction minu(Fraction f1, Fraction f2){
    Fraction result;
    result.up = f1.up * f2.down - f1.down * f2.up;
    result.down = f1.down * f2.down;
    return reduction(result);
}

//4.分数乘法
Fraction multi(Fraction f1, Fraction f2){
    Fraction result;
    result.up = f1.up * f2.up;
    result.down = f1.down * f2.down;
    return reduction(result);
}

//5.分数除法
Fraction divide(Fraction f1, Fraction f2){
    Fraction result;
    result.up = f1.up * f2.down;
    result.down = f1.down * f2.up;
    return reduction(result);
}

//分数的输出
void showResult(Fraction r){
    r = reduction(r);
    if(r.down == 1) printf("%lld", r.up);
    else if(abs(r.up) > r.down){
        printf("%d %d/%d", r.up / r.down, abs(r.up) % r.down, r.down);
    }else{
        printf("%d/%d", r.up, r.down);
    }
}

四、素数:是指除了1和本身之外,不能被其他数整除的一类数;1既不是素数,也不是合数;

//1.素数的判断
bool isPrime(int n){
    if(n <= 1) return false;
    //int sqr = (int)sqrt(1.0 * n);
    for(int i = 2; i <= n; i++){
        if(n % i == 0) return false;
    }
    return true;
}

//2.素数表的获取
const int maxn = 101;//表长
int prime[maxn], pNum = 0;//prime数组存放所有素数,pNum为素数个数
bool p[maxn] = {0};//p[i] == true表示i是素数
void Find_Prime(){
    for(int i = 0; i < maxn; i++){
        if(isPrime(i) == true){
            prime[pNum++] = i;
            p[i] = true;
        }
    }
}


void Find_Prime(){
    for(int i = 2; i < maxn; i++){
        if(p[i] == false){
            prime[pNum++] = i;
            for(int j = i + i; j < maxn; j += i){
                p[j] = true;
            }
        }
    }
}

五、质因子分解

#include<iostream>
#include<cmath>
using namespace std;
const int maxn = 100010;
bool isPrime(int n){
	if(n <= 1) return false;
	int sqr = (int)sqrt(1.0 * n);
	for(int i = 2; i <= sqr; i++){
		if(n % i == 0) return false;
	}
	return true;
} 
int prime[maxn], pNum = 0;
void Find_Prime(){
	for(int i = 1; i < maxn; i++){
		if(isPrime(i) == true){
			prime[pNum++] = i;
		}
	}
}
struct factor{
	int x, cnt;
}fac[10];
int main(){
	Find_Prime();
	int n, num = 0;
	scanf("%d", &n);
	if(n == 1) printf("1=1");
	else{
		printf("%d=", n);
		int sqr = (int)sqrt(1.0*n);
//
		for(int i = 0; i < pNum && prime[i] <= sqr; i++){
			if(n % prime[i] == 0){
				fac[num].x = prime[i];
				fac[num].cnt = 0;
				while(n % prime[i] == 0){
					fac[num].cnt++;
					n /= prime[i];
				}
				num++;
			}
			if(n == 1) break;
		}
		if(n != 1){
			fac[num].x = n;
			fac[num++].cnt = 1;
		}
		for(int i = 0; i < num; i++){
			if(i > 0) printf("*");
			printf("%d", fac[i].x);
			if(fac[i].cnt > 1){
				printf("^%d", fac[i].cnt);
			}
		}
	}
//
	return 0;
}

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转载自www.cnblogs.com/tsruixi/p/13200010.html
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