//利息计算
//启动资金10000
//固定利息 = 本金*0.1%
//复合利息 = (本金+产生的利息)*0.05%
#include<stdio.h>
int main()
{
float a = 10000;
float b = 10000;
int count = 0;
float ra = a * 0.1;
do
{
a += ra; //固定利息
b += (b * 0.05); //复利
count++;
} while (b < a);
printf("%d 年后, 黑夜的投资额超过小甲鱼！\n", count);
printf("小甲鱼的投资额是: %.2f\n", a);
printf("黑夜的投资额是: %.2f\n", b);
return 0;
}
//本金400万 每年0.08利息 每年开支50万 算几年后破产
#include<stdio.h>
int main()
{
float a = 4000000;
int count = 0;
while (a > 0)
{
a -= 500000; //开销
a += a * 0.08; //利息
count++;
}
printf("%d年后，小甲鱼败光了所有的家产，再次回到一贫如洗......", count);
return 0;
}
//计算 (pi) 的值
#include<stdio.h>
int main()
{
double a = 0;
int flag = 1;
// (pi/4) = 1 - (1/3) + (1/5) - ...... + flag(1/n+2)
for (int i = 1; i < 100000000; i += 2)
{
a += 1.0 * flag / i;
flag = -flag;
}
a *= 4.0;
printf("Pi = %.7lf", a);
return 0;
}
//兔子出生2个月后 每月可以生一对兔子 2年后(24个月)有多少兔子
//斐波那契数列
#include<stdio.h>
int main()
{
int f1 = 0;
int f2 = 1;
int f3 = 0;
int i = 0;
int count = 2;
printf("%d对(1月)\t", f2);
for (i = 2; i <= 24; i++, count++)
{
f3 = f1 + f2;
printf("%d对(%d月)\t", f3, count);
//后移一位, 为下次循环做准备
f1 = f2;
f2 = f3;
//换行 格式
if (count % 6 == 0)
printf("\n");
}
return 0;
}
{:10_277:}
测试题:
0:10个,在内循环中的循环条件是j<10时循环,而最后j=10时内层循环退出,而外层循环的条件是需要j!=10,而此时j=10,外层循环也退出了.所以将打印10个'A'
1:0个,因为i++是先取出变量i的值,然后在自增1,那么第一次循环是while内的值是0,循环直接不执行.
2: a, b, c,都是lvalue.
3:a=14, b=5, c=9
4:z = x < 0 ? -x : x
5:
A:
if (size > 12)
{
cost = cost * 1.05;
flag = 2;
}
bill = cost * flag;
B:
if (ibex > 14)
{
sheds = 3;
}else
{
sheds = 2;
}
help = 2 * sheds;
c:
scanf("%d", &score);
while(score < 0)
{
count ++
scanf("%d", &score);
}
printf("count = %d\n", count);
动动手:
0:
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
float xjysmoey = 10000, hysmoney = 10000;
int years = 1;
while (xjysmoey >= hysmoney) {
xjysmoey = 10000 + 10000 * years * 0.1;
hysmoney = hysmoney + hysmoney * 0.05;
years++;
}
printf("%d年后,黑夜的投资额超过小甲鱼!\n小甲鱼的投资额是:%."
"2f\n黑夜的投资额是:%.2f\n",
years - 1, xjysmoey, hysmoney);
return EXIT_SUCCESS;
}
1:
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
float money = 4000000;
int years = 0;
while (money >= 0) {
money -= 500000;
money += money * 0.08;
years++;
}
printf("%d年之后,小甲鱼败光了所有的家产,再次回到一贫如洗......\n", years);
return EXIT_SUCCESS;
}
2:#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
double PI = 0.0, n = 1, term = 1.0; // PI 是圆周率n是分母,term是该项的值;
int sign = 1; // sign代表正负值;
while (fabs(term) >= 1e-8) { // 1e-8 就是10 ^ (-8) 在程序中的表达式.
PI = PI + term;
sign = -sign;
n = n + 2;
term = sign / n;
}
PI = PI * 4;
printf("PI的值为:%15.7f\n", PI);
return EXIT_SUCCESS;
}
3:
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
long a;
int month = 3;
int firstmonth, sencondmonth;
firstmonth = 1;
sencondmonth = 1;
while (month <= 24) {
a = firstmonth + sencondmonth;
sencondmonth = firstmonth;
firstmonth = a;
month++;
}
printf("%ld\n", a);
return EXIT_SUCCESS;
}
10个A
尽力了
本帖最后由 小正阿 于 2026-6-4 21:01 编辑
0. 10 个。
1. 0个。
2. 不清楚。
3.
a = 14
b = 5
c = 9
111
111
+1
看
.
.