acm问题解答1.3

We need to find out the number of numbers with the following properties (including the natural number n of input). First, input a natural number n (n<=1000), and then deal with the natural number according to the following method: 1. do not do any treatment; 2. add a natural number to its left, but the natural number can not exceed half of the highest number of the original number; after 3., the number will continue to be processed until the natural number can not be added.

[input] the first row is a number T, which means that each group of data has one row after each T group data, and each row has a number of n.

[output] each group of data takes one line, one number, to represent the number of the number of conditions that satisfy the condition.

[input case] 16

[output example] 6

solved:

 #include <stdio.h>

void main()
 {
int i,c,a[1000],n,z,sum[1000],j,temp,m;
    sum[0] = 0,sum [1] = 1;
printf("请输入数据组数：");
scanf("%d",&c);
    printf("请输入数据：");
for(j=0;j<c;j++){
scanf("%d",&a[j]);
m=a[j];
    n=a[j];
 while(n>0)  // 当n=0或小于0时，退出循环，此时所以位数都已取出
    {
        temp = n % 10;   
        n /= 10;  // 将去掉n的最低位，次低位变为最低位
    }
if(temp==1) printf("%d\n",temp);
else{
    for(i=2 ; i <= m; i++)
    {
       z = sum[i/2] + 1;
   sum [i] = sum[i-1] + z;}
    printf("%d\n",sum[m] - sum[m-1]);
}}
}