zoukankan      html  css  js  c++  java
  • uvaoj1225Digit Counting(暴力)

    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

     题意:把1到n的数字排列起来,数0到9之间的数出现的次数,最后输出每个出现的次数

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int ans[15];
     4 int main()
     5 {
     6     int t;
     7     scanf("%d",&t);
     8     while(t--)
     9     {
    10         memset(ans,0,sizeof(ans));
    11         int n;
    12         scanf("%d",&n);
    13         for(int i=1;i<=n;i++)
    14         {
    15             int temp=i;
    16             while(temp>0)
    17             {
    18                 ans[temp%10]++;
    19                 temp/=10;
    20             }    
    21         }
    22         for(int i=0;i<9;i++)
    23         {
    24             printf("%d ",ans[i]);
    25         }
    26         printf("%d
    ",ans[9]);
    27     }
    28     return 0;
    29 }
  • 相关阅读:
    HDU 5087 (线性DP+次大LIS)
    POJ 1064 (二分)
    Codeforces 176B (线性DP+字符串)
    POJ 3352 (边双连通分量)
    Codeforces 55D (数位DP+离散化+数论)
    POJ 2117 (割点+连通分量)
    POJ 1523 (割点+连通分量)
    POJ 3661 (线性DP)
    POJ 2955 (区间DP)
    LightOJ 1422 (区间DP)
  • 原文地址:https://www.cnblogs.com/fqfzs/p/9863204.html
Copyright © 2011-2022 走看看