第4章 程序段的反复执行4 多重循环练习（题及答案）

（1）程序阅读

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main(){int i,j,n;cin >> n;for(i = 1; i <= n; i++){for(j = 1; j <= n - i;j++)cout << " ";for(j = 1; j < i; j++)cout << "*";cout << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main(){int i,j,n;cin >> n;for(i = 2; i <= n; i++){j = i - 1;while(j > 1 && i % j != 0)j--;cout << i << "(" << j << ")\n";}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int i, m, n = 0;for(i = 1; i <= 5; i++) {m = i % 2;while(m-- > 0) n++;}cout << m << "," << n;return 0;
}
-1,3

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;cout << n << "=";for(int i = 2; i <= n; i++) {for(; n % i == 0;) {n = n / i;cout << i;if(n != 1) cout << "*";}}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;assert(1 <= n && n <= 20);for (int row = 1; row <= n; row++) {for (int col = 1; col <= n + row - 1; col++) {if (col <= n - row) { cout << " ";} else { cout << "*";}}cout << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;assert(1 <= n && n <= 10);for(int row = 1; row <= n; row++) {int i = 1;for(int col = 1; col <= n + row - 1; col++) {if(col < n - row + 1) {cout << " ";} else {cout << i++;}}cout << endl;}for(int row = n - 1; row >= 1; row--) {int i = 1;for(int col = 1; col <= n + row - 1; col++) {if(col < n - row + 1) {cout << " ";} else {cout << i++;}}cout << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;assert(1 <= n && n <= 9);for(int row = 1; row <= n; row++) {for(int col = 1; col <= row; col++) {cout << col << "*" << row << "=" << col*row << " ";}cout << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;assert(100 <= n);int ways = 0;for(int i = 0; i <= n / 50; i++) {ways += (n - i * 50) / 20 + 1;}cout << ways << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;assert(n >= 1);int sumOfDigits = 0;while(1) {while(n > 0) {sumOfDigits += n % 10;n /= 10;}if (sumOfDigits < 10) {break;} else {n = sumOfDigits;sumOfDigits = 0;}}cout << sumOfDigits << endl;return 0;
}

典型的数论题目，考查的是最大公约数（gcd）与最小公倍数（lcm）的定义和性质。

#include <iostream>
using namespace std;
int main() {int x0, y0;cin >> x0 >> y0;if (y0 % x0 != 0) {cout << 0 << endl; // 如果不能整除，直接输出0return 0;}int k = y0 / x0;int count = 0;for (int a = 1; a * a <= k; ++a) {if (k % a == 0) {int b = k / a;// 计算 a 和 b 的最大公约数（不用函数）int m = a, n = b;while (n != 0) {int r = m % n;m = n;n = r;}int d = m; // 此时 d = gcd(a, b)if (d == 1) {if (a == b)count += 1; // (a, a) 只算一种elsecount += 2; // (a, b) 和 (b, a) 算两种}}}cout << count << endl;return 0;
}