题目如图,这个不算很难,但是想看一下程序设计的思路
,谢谢啦
第4题:
#include <stdio.h>
int numChain(int n)
{
int t;
printf("%d",n);
while(n>=10){
t = 0;
while(n>0){
t += (n%10)*(n%10);
n /= 10;
}
n = t;
printf("->%d",n);
}
printf("\n");
return n;
}
int main()
{
int i, N, count = 0;
scanf("%d",&N);
for (i=N;i>0;i--){
if (numChain(i)==1) count++;
}
printf("其中,数字链尾数是1的个数:%d",count);
return 0;
}
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main()
{
char s[100] = {0};
int n=0,i,j=0;
scanf("%s", s);
for (i = strlen(s)-1; i >= 0; i--)
n += (s[i]-'0') * pow(2,j++);
printf("%d\n",n);
return 0;
}