mpi
发表于 2020-10-21 08:36:49
nice
夏空彼方
发表于 2020-10-21 11:10:42
11
csgo
发表于 2020-10-21 15:50:31
恢复
鱼跃此时
发表于 2020-10-21 16:51:52
WHY
jxshop
发表于 2020-10-21 17:17:09
支持甲鱼
Blueboy
发表于 2020-10-21 19:18:45
1
goingkui
发表于 2020-10-21 21:52:03
0)10个A
1)11个B
2)c,b,a
3)13,5,10
4)z=(x>=0?x:-x)
5)
A.if(size>12)
{
cost=cost*1.05;
flag=2;
}
else if
{
bill=cost*flag;
}
B.if(ibex>14)
{
sheds=3;
}
else if()
{
sheds=2;
help=2*sheds;
}
C. do
{
scanf("%d",&score);
pritnf("count=%d\n",count);
count++;
}while(score<0);
0)
#include <stdio.h>
#include<math.h>
int main()
{
float a,b;
a=b=10000.0;
int i;
for(i=1;a>=b;i++)
{
a=10000.0*0.1*i+10000.0;
b=10000.0*(pow(1.05,i));
}
printf("%d年后,投资额超过!\n鱼的投资额:%.2f\n夜的投资额:%.2f",i,a,b);
return 0;
}
1)#include <stdio.h>
#include<math.h>
int main()
{
float a;
a=400.0;
int i;
for(i=1;a>0;i++)
{
a=(a-50)*(1+0.08);
}
printf("%d年后,破产",i-1);
return 0;
}
2)
#include <stdio.h>
#include<math.h>
int main()
{
double a,b,c;
int i;
a=1;
b=0;
c=pow(10,-16);
for(i=1;a*a>c;i++)
{
a=(pow(-1,i+1))/(2*i-1);
b=b+a;
}
printf("%.8lf",b);
return 0;
}
学会再改名
发表于 2020-10-21 22:09:46
难受
拾叁。
发表于 2020-10-21 22:55:58
1
18955994055
发表于 2020-10-22 10:51:18
哈哈哈哈
reol0316
发表于 2020-10-22 14:30:04
看看答案
十三昆
发表于 2020-10-22 16:58:37
题目挺好
OYSCAR
发表于 2020-10-23 09:17:42
l
Astronaut丶
发表于 2020-10-23 14:45:02
1
吻你不厌
发表于 2020-10-23 19:37:58
90
11
a,b,c
14 3 9
z = x >=0? x : -x;
机智如莪
发表于 2020-10-23 22:33:50
0. 10
1. 0
2. 5
3.
a = 14
b = 5
c = 9
4.
#include <stdio.h>
int main()
{
int x, z;
scanf("%d", &x);
z = x > 0 ? x : -x;
printf("%d", z);
return 0;
}
5.
A.
if (size > 12)
{
cost *= 1.05;
flag = 2;
}
bill = cost * flag;
B.
if (ibex > 14)
{
sheds = 3;
}
sheds = 2;
help = 2 * sheds;
C.
while (1)
{
readin: scanf("%d", &score);
if (score < 0)
{
printf("count = %d\n", count);
}
count++;
}
0.
#include <stdio.h>
int main()
{
int time, a = 10000, b = 10000;
double sa = a, sb = b;
for (b, sb; sb <= sa; time++)
{
sa += a * 0.1;
sb += sb * 0.05;
}
printf("%d年后超过\nA投资额是%.2f\nB投资额是%.2f", time, sa, sb);
}
1.
#include <stdio.h>
int main()
{
double a = 400;
int year;
for (a; a > 0; year++)
{
a = a * 1.08 - 50;
}
printf("%d年后没钱", year);
return 0;
}
2.
#include <stdio.h>
#include <math.h>
int main()
{
double PI, a = 1.0000000;
int i = 1;
for (PI, a, i; fabs(a) > 1.0E-8;)
{
a = 1.0000000 / i;
PI += a;
i = -(i + 2 );
}
printf("Π的值约等于%.7f", PI);
return 0;
}
3.
#include <stdio.h>
int main()
{
int b = 2, t, i, s = 0;
for (b, s, t; i < 24; i++)
{
t = b + s;
b += s;
s = t;
}
printf("兔子数量为%d", b + s);
return 0;
}
打码到秃头
发表于 2020-10-24 15:55:18
.
3120586067
发表于 2020-10-24 18:06:00
谢谢大佬
mbyul
发表于 2020-10-25 18:46:35
1
kiseki1223
发表于 2020-10-25 19:29:38
hhhhhhhhhhhhhh