磊
1
0.10
1.0
2.a b c
3.13 3 8
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;
}
sheds = 2;
help = 2 * sheds;
C.do{
readin: scanf("%d", &score);
printf("count = %d\n", count);
count++;
}while(score < 0)
0.#include <stdio.h>
int main()
{
float xjy=10000,hy=10000;
int i;
do
{
xjy+=1000;
hy*=1.05;
i++;
}while(hy<xjy);
printf("%d年后,黑夜的投资额超过小甲鱼!\n",i);
printf("小甲鱼的投资额是:%.2f\n",xjy);
printf("黑夜的投资额是:%.2f\n",hy);
return 0;
}
1.#include <stdio.h>
int main()
{
int s=400,i=0;
do
{
s-=50;
s*=1.08;
i++;
}while(s>0);
printf("%d年之后,小甲鱼败光了所有的家产,再次回到一贫如洗......",i);
}
2.#include <stdio.h>
int main()
{
double pi=0;
double fz=1,fm=1,s=0;
int sign=1;
do
{
s+=(fz*sign)/fm;
fm+=2;
sign=-sign;
}while((fz/fm)>1e-8);
pi=4*s;
printf("%.7f",pi);
return 0;
}
3.#include <stdio.h>
int main()
{
int s1=1,s2=1,i=0;
do
{
s1+=s2;
s2+=s1;
i++;
}while(i<11);
printf("%d",s2);
return 0;
}
答案
10
10
c = 5, b = c, a = b
14 5 9
if(x < 0) { z = -x; } else if (x > 0) { z = x; }
A:
if (size > 12) {cost = cost * 1.05; flag = 2;};
else { bill = cost * flag};
B:
if (ibex > 14) { sheds = 3}; else { sheds = 2; help = 2 * sheds;}
C:
if (score < 0) { printf("count = %d\n", count)} else {count++}
#include <stdio.h>
#include <math.h>
int main(){
int count = 1, TF = 1;
double x, y;
do {
x = 10000 * 0.1 * count + 10000;
y = 10000 * pow((1 + 0.05), count);
if (x < y){
TF = 0;
}
count++;
} while (TF);
printf("%d年后,黑夜的投资额超过小甲鱼!\n", count - 1);
printf("小甲鱼的投资额是:%.2f\n", x);
printf("黑夜的投资额是:%.2f\n", y);
return 0;
}
#include <stdio.h>
#include <math.h>
int main(){
double base = 4000000.0;
int count = 1;
do {
base = (base - 500000.0) * (1 + 0.08);
count++;
} while (base >= 0);
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", count - 1);
return 0;
}
#include <stdio.h>
#include <math.h>
int main(){
int count=0, num = 1, i=1;
double sum=0.0;
double term;
do {
term = (double)1 / num;
sum += i * term;
i = -i;
num += 2;
} while (fabs(term) >= pow(10, -8));
printf("Pi的近似值为%.7lf", sum * 4);
return 0;
}
#include <stdio.h>
#include <math.h>
int main(){
int parent = 2, child1 = 0, child2 = 0;
int time;
for (time = 0; time < 24; time++)
{
parent += child2;
child2 = child1;
child1 = parent / 2;
}
printf("两年后可以繁殖出%d对", child1 / 2);
return 0;
}
看答案
daan
1
0.10个
1.11个
2.a b c
3.a=14 b=3 c=10
4.z = x>0?x:-x
5.if (size > 12)
{
cost = cost * 1.05;
flag = 2;
continue;
}
bill = cost * flag;
B
if (ibex > 14)
{
sheds = 3;
continue;
}
sheds = 2;
help = 2 * sheds;
C
if (score < 0)
{
printf("count = %d\n", count);
continue;
}
count++;
scanf("%d", &score);
{:5_101:}
1