第4章 程序段的反复执行3 do-whiile语句P139练习（题及答案）

(1)程序阅读

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int n;cin >> n;do {cout << n % 2;n /= 2;} while(n != 0);return 0;
}
输入4   输出：001
输入0   输出：0

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int i, n;cin >> n;i = n - 1;doi--;while(i > 1 && n % i != 0);cout << i;return 0;
}
输入100     输出：50
输入79      输出：1
输入2       输出：0

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int i = 1, s = 3;do {s += i++;if(s % 7 != 0) ++i;} while(s < 15);cout << i;return 0;
}
输出8

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int s = 1;                   // sign用来表示数值的符号double n = 1, pi = 0, t = 1; // pi开始代表多项式的值，最后代表π的值，n代表分母，// term代表当前项的值while (fabs(t) > 1e-6) { //检查当前项term的绝对值是否大于10的（-6）次方pi = pi + t; //当前项t累加到pi中n = n + 2;   // n+2是下一项的分母s = -s;      // s代表符号，与上一项相反的符号t = s / n;   //求出下一项的值}pi = pi * 4;               //π的值printf("pi=%10.6f\n", pi); //输出π的近似值return 0;
}

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {int count = 1;float hight, dis;hight = 200;dis = 200;do {hight = hight / 2;dis += 2 * hight;count++;} while(hight >= 0.5);cout << "小球弹跳的总路程是:" << dis << endl;return 0;
}
//599.219

#include <bits/stdc++.h>
using namespace std;
int main() {char a[10];int b[10], lena;scanf("%s", a);lena = strlen(a);if (strchr(a, '-')) {printf("-");a[0] = '0';for (int i = 0; i < lena; i++) {a[i] = a[i + 1];}lena--;}for (int i = 0; i < lena; i++) {b[i] = a[i] - 48;}for (int i = lena - 1; i >= 0; i--) {if(b[i] != 0)printf("%d", b[i]);}return 0;
}

1/4

#include <iostream>
#include <algorithm>
using namespace std;
int main() {int n, j = 0, i = 0;cin >> n;while(n > j) {i++;j += i;}if(i % 2 != 0) {cout << j + 1 - n << '/' << i + n - j << endl;} elsecout << i + n - j << '/' << j + 1 - n << endl;return 0;
}

输出是8，不是9。

#include <bits/stdc++.h>
using namespace std;
//汤永红
int main() {long long x, maxy = 0;cin >> x;for(int i = x; i <= 2000000; i++) {maxy += 1;if(x % 2 == 0) {x = x / 2;if(x == 1) {cout << maxy;break;}} else {x = x * 3 + 1;}}return 0;
}