我正在做的任务是输入5个具有5位数字的随机数,以对其进行压缩并求和。那么我们如何压缩它们呢?我们摆脱了第二和第四位数字。例如12345到10305。
这是我的代码。
int main()
{
int n=5,i,j,number5,num1,num3,numb5,sum;
for(i=0;i<n;i++) // 5 times we read next for loop right ?
for(j=0;j<i;j++){ // this loop read 5 times 5 digits number
scanf("%d",&number5); // scanf 1 number
while(number5){ // while number isn't 0 ( false )
num1=number5/10000;
num1*=10000;
num3=(number5%10000)%1000/100;
num3*=100;
numb5=(number5%10000)%1000%100%10;
numb5*=1;// mathematic operations to get to 1st third and 5th number
number5=0; // set the number5 to 0 so we can go out of while right ?
}
sum=num1+num3+numb5; // we get the sum of the first 5 digits and we get it on the second when j++ right ?
}
printf("%d",sum);// on the end of all five number with 5 digits we get the sum right ?
}
那我的for
循环为什么只运行两次而不是五次?
答案 0 :(得分:0)
答案 1 :(得分:0)
只需在第二个循环中将j<i
移至j<5
,您就可以得到五个{5}压缩数字的总和,n
次。
答案 2 :(得分:0)
我在摘要中提到的错误是唯一的问题,其他一切都很好。 请参阅附件。
// only two loops are required
for(int i=0;i<5;i++){
scanf("%d",&number5);
sum=0;
while(number5){
num1=number5/10000;
num1*=100;//it should be 100 isntead of 10000
num3=(number5%10000)%1000/100;
num3*=10;// it should be 10 istead of 100
numb5=(number5%10000)%1000%100%10;
numb5*=1;
number5=0;
}
sum=num1+num3+numb5;
printf("%d\n",sum);
}
答案 3 :(得分:0)
#include <stdio.h>
#include <stdlib.h>
int main()
{
int n=5,i,j,number5,num1,num3,numb5;
int sum=0;
for(i=0;i<n;i++) {
scanf("%d",&number5);//runs 5 times to get 5 inputs
sum+=(number5/10000)*10000; // take the value of first position
sum+=((number5/100)%10)*100; // take the value of third position
sum+=(number5%10);// take the value of fifth position
}
printf("%d",sum);
}