Digit Counting Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu Submit Status Description Download as PDF Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N(1 < N < 10000) . After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, with N = 13 , the sequence is: 12345678910111213 In this sequence, 0 appears once, 1 appears 6 times, 2 appears 2 times, 3 appears 3 times, and each digit from 4 to 9 appears once. After playing for a while, Trung gets bored again. He now wants to write a program to do this for him. Your task is to help him with writing this program. Input The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets. For each test case, there is one single line containing the number N . Output For each test case, write sequentially in one line the number of digit 0, 1,...9 separated by a space. Sample Input 2 3 13 Sample Output 0 1 1 1 0 0 0 0 0 0 1 6 2 2 1 1 1 1 1 1
给你一个数字 从1 开始到这个数字 意见 一共出现了 多少个 1 - 9 统计一下 n<10000
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 char str[104]; 7 int main() 8 { 9 int n; 10 while (scanf("%d",&n)!=EOF) 11 { 12 while (n--) 13 { 14 scanf("%s",str); 15 int len = strlen(str); 16 for (int k,i = 1 ; i <= len ; ++ i) //字符的长度 i是 循环 的长度 17 { 18 if (len%i == 0) // 如果是循环的话 那么一定是 可以整除的. 19 { 20 for (k = i ; k < len ; ++ k) //从第二个字符开始比较 21 if (str[k] != str[k%i]) // 不相等的话跳出去 //如果确定了字符串长度 那么 可以一直贴合到 k=len 22 break; 23 } 24 if (k == len) //知道最后 都不知道 循环长度 那么 字符长度就是 循环长度 25 { 26 printf("%d ",i); 27 break; 28 } 29 } 30 if (n) 31 printf(" "); 32 } 33 } 34 return 0; 35 }