31.计算1-2+3-4+5-...+99-100=?
#include <stdio.h>
#include <math.h>
int main() {
int sum = 0;
for (int i = 1; i <= 100; i++) {
sum += pow(-1, i - 1) * i;
}
printf("sum=%d", sum);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=-50
Process finished with exit code 0
32.计算1/1+1/2+1/3+...1/100=?
#include <stdio.h>
#include <math.h>
int main() {
float sum = 0;
for (int i = 1; i <= 100; i++) {
sum += pow(i,-1);
}
printf("sum=%f", sum);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=5.187378
Process finished with exit code 0
33.计算1/1+1/2+1/3+...+1/n=?,n=10^6
#include <stdio.h>
#include <math.h>
int main() {
float sum = 0;
for (int i = 1; i <= 1e6; i++) {
sum += pow(i, -1);
}
printf("sum=%f", sum);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=14.357358
Process finished with exit code 0
34.输入两个整数m和n，求其最大公约数和最小公倍数
#include <stdio.h>
#include <math.h>
int main() {
int m, n, r, p;
printf("input m,n: ");
scanf("%d,%d", &m, &n);
p = m * n;
do {
r = m % n;
m = n;
n = r;
} while (r != 0);
printf("最大公约数:%d\n", m);
printf("最小公倍数:%d", p / m);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
input m,n: 6,8
最大公约数:2
最小公倍数:24
Process finished with exit code 0
35.计算 2+22+222+2222+22222=？(此时n=5)
#include <stdio.h>
int main() {
int n, s = 0, t = 0;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
t = t * 10 + 2;
s += t;
}
printf("%d\n", s);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
5
24690
Process finished with exit code 0
36.计算 2/1+3/2+5/3+8/5+... 求前20项
#include <stdio.h>
int main() {
float m = 2, n = 1, t, s = 0;
for (int i = 1; i <= 20; i++) {
s += m / n;
t = m;
m += n;
n = t;
}
printf("sum=%f\n", s);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=32.660263
Process finished with exit code 0
37.计算 1!+2!+3!+...+20!=?
#include <stdio.h>
int main() {
float s = 0, f = 1;
for (int i = 1; i <= 20; i++) {
f *= i;
s += f;
}
printf("sum=%f\n", s);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=2561327455189073920.000000
Process finished with exit code 0
38.计算
\sum_{k=1}^{100}{k}+\sum_{k=1}^{50}{k^2}+\sum_{k=1}^{10}\frac{1}{k}=?
#include <stdio.h>
int main() {
int s1 = 0, s2 = 0;
float s3 = 0;
for (int i = 1; i <= 100; i++)
s1 += i;
for (int i = 1; i <= 50; i++)
s2 += i * i;
for (int i = 1; i <= 10; i++)
s3 += 1.0 / i;
printf("%f", s1 + s2 + s3);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
47977.929688
Process finished with exit code 0
39.水仙花数：一个3位数，各位数字的立方和等于它本身（如：153）,打印全部水仙花数
1).
#include <stdio.h>
#include <math.h>
int main() {
int x, y, z, s;
for (int m = 100; m < 1000; m++) {
x = m / 100;
y = m / 10 % 10;
z = m % 10;
s = pow(x, 3) + pow(y, 3) + pow(z, 3);
if (m == s)
printf("%d is a water flower number.\n", m);
}
return 0;
}
2).
#include <stdio.h>
#include <math.h>
int main() {
int x, s;
for (int m = 1; m <= 9; m++)
for (int i = 0; i <= 9; i++)
for (int j = 0; j <= 9; j++) {
x = m * 100 + i * 10 + j;
s = pow(m, 3) + pow(i, 3) + pow(j, 3);
if (x == s)
printf("%d is a water flower number.\n", x);
}
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
153 is a water flower number.
370 is a water flower number.
371 is a water flower number.
407 is a water flower number.
Process finished with exit code 0
40.如果一个数恰好等于它的因子之和，这个数为“完数”（例如:6=1+2+3），找出1000以内的完数，按照:“6 its factors are 1 2 3” 格式打印
#include <stdio.h>
#include <math.h>
int main() {
int s;
for (int i = 1; i <= 1000; i++) {
s = 0;
for (int j = 1; j <= i / 2; j++)
if (i % j == 0)
s += j;
if (s == i) {
printf("%d it's factors are ", i);
for (int j = 1; j <= i / 2; j++) {
if (i % j == 0)
printf("%d ", j);
}
printf("\n");
}
}
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
6 its factors are 1 2 3
28 its factors are 1 2 4 7 14
496 its factors are 1 2 4 8 16 31 62 124 248
Process finished with exit code 0
41.一个球从100m的高度落下，每次落地后反跳回原高度的一半，再落下，再反弹。求它第10次落地时，共经过多少米，第10次反弹多高
#include <stdio.h>
#include <math.h>
int main() {
float sum = 100, high = 50;
for (int i = 0; i < 10; i++) {
sum += 2 * high;
high /= 2;
}
printf("sum=%.2f,high=%f\n", sum, high*2);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
sum=299.90,high=0.024414
Process finished with exit code 0
42.（谭浩强c语言课本P141）猴子吃桃
#include <stdio.h>
#include <math.h>
int main() {
int x1, x2 = 1;
for (int day = 9; day >= 1; day--) {
x1 = 2 * (x2 + 1);
x2 = x1;
}
printf("num=%d\n", x1);
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
num=1534
Process finished with exit code 0
43.每个月兔子的总和依次为1，1，2，3，5，8，13...这就是斐波那契数列
{
f1=1(n=1);
f2=1(n=2);
fn=f(n-1)+f(n-2)(n>=3);
}
#include <stdio.h>
#include <math.h>
int main() {
int f1 = 1, f2 = 1, f;
printf("%5d %5d", f1, f2);
for (int i = 3; i <= 20; i++) {
f = f1 + f2;
printf("%5d ", f);
if (i % 5 == 0)
printf("\n");
f1 = f2;
f2 = f;
}
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
1 1 2 3 5
8 13 21 34 55
89 144 233 377 610
987 1597 2584 4181 6765
Process finished with exit code 0
44.判断一个数是否为素数
#include <stdio.h>
#include <math.h>
int main() {
int m, i;
scanf("%d", &m);
for (i = 2; i <= m / 2; i++)
if (m % i == 0)
break;
if (m/2 == i-1)
printf("Yes");
else
printf("No");
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
5
Yes
Process finished with exit code 0
45.成绩平均值
#include <stdio.h>
#include <math.h>
int main() {
int sum = 0, s[10];
float ave;
for (int i = 0; i <= 9; i++) {
scanf("%d", &s[i]);
sum += s[i];
}
ave = sum / 10.0;
printf("ave=%f\n", ave);
for (int j = 0; j < 10; j++) {
printf("%d", s[j]);
}
return 0;
}
输出：
/Users/wuyanbo/Documents/ClionProject/cmake-build-debug/ClionProject
1
2
3
4
5
6
7
8
9
0
ave=4.500000
1234567890
Process finished with exit code 0
