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  • UVa 1225 / UVALive 3996 Digit Counting 数数字(字符统计)

    Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

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    Description

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    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

    大致题意:给出一个数n,把从1到n按顺序写在一起,统计数字0~9各个数出现的次数。
     1 #include<iostream>
     2 #include<string.h>
     3 #include<cstdio>
     4 using namespace std;
     5 int main()
     6 {
     7     int T,b,c,i,n,j,e,f;
     8     scanf("%d",&T);
     9     while(T--)
    10     {
    11         int k[10]={0};
    12 
    13         scanf("%d",&n);
    14         for(i=1;i<=n;i++)
    15         {
    16             f=i;
    17             j=0;
    18             int a[5]={0};
    19             while(f!=0)
    20             {
    21                 a[j++]=f%10;
    22                 f/=10;
    23             }
    24             for(int e=0;e<j;e++)
    25                 k[a[e]]++;
    26         }
    27         for(i=0;i<10;i++)
    28         {
    29             if(i==0)
    30                 printf("%d",k[i]);
    31             else
    32                 printf(" %d",k[i]);
    33         }
    34         printf("
    ");
    35     }
    36     return 0;
    37 }
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  • 原文地址:https://www.cnblogs.com/caterpillarofharvard/p/4232099.html
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