• POJ 2282 The Counting Problem


    The Counting Problem
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 4612   Accepted: 2366

    Description

    Given two integers a and b, we write the numbers between a and b, inclusive, in a list. Your task is to calculate the number of occurrences of each digit. For example, if a = 1024 and b = 1032, the list will be
    1024 1025 1026 1027 1028 1029 1030 1031 1032

    there are ten 0's in the list, ten 1's, seven 2's, three 3's, and etc.

    Input

    The input consists of up to 500 lines. Each line contains two numbers a and b where 0 < a, b < 100000000. The input is terminated by a line `0 0', which is not considered as part of the input.

    Output

    For each pair of input, output a line containing ten numbers separated by single spaces. The first number is the number of occurrences of the digit 0, the second is the number of occurrences of the digit 1, etc.

    Sample Input

    1 10
    44 497
    346 542
    1199 1748
    1496 1403
    1004 503
    1714 190
    1317 854
    1976 494
    1001 1960
    0 0
    

    Sample Output

    1 2 1 1 1 1 1 1 1 1
    85 185 185 185 190 96 96 96 95 93
    40 40 40 93 136 82 40 40 40 40
    115 666 215 215 214 205 205 154 105 106
    16 113 19 20 114 20 20 19 19 16
    107 105 100 101 101 197 200 200 200 200
    413 1133 503 503 503 502 502 417 402 412
    196 512 186 104 87 93 97 97 142 196
    398 1375 398 398 405 499 499 495 488 471
    294 1256 296 296 296 296 287 286 286 247
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <iomanip>
    #include <cmath>
    #include <ctime>
    #include <map>
    #include <set>
    using namespace std;
    #define lowbit(x) (x&(-x))
    #define max(x,y) (x>y?x:y)
    #define min(x,y) (x<y?x:y)
    #define MAX 100000000000000000
    #define MOD 1000000007
    #define pi acos(-1.0)
    #define ei exp(1)
    #define PI 3.141592653589793238462
    #define INF 0x3f3f3f3f3f
    #define mem(a) (memset(a,0,sizeof(a)))
    typedef long long ll;
    int a[10],b[10],n,m;
    //以1234567为例子
    void dfs(int n,int *a,int len)
    {
        if(n<=0) return ;
        int ans=n%10;
        int pos=n/10;
        for(int i=1;i<=ans;i++)//开始时len为1,即dfs到个位(1~7)各出现了一次
            a[i]+=len;
        while(pos>0)
        {
            a[pos%10]+=(ans+1)*len;//1~7累加时,前面也相应重复出现并且(ans+1)是0的情况
            pos/=10;
        }
        for(int i=0;i<=9;i++)
            a[i]+=(n/10)*len;//从1~123456,需要0~9的个数
        dfs(n/10-1,a,len*10);
    }
    int main()
    {
        while(scanf("%d%d",&n,&m) &&(n && m))
        {
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            if(n>m) swap(n,m);
            n--;
            dfs(m,a,1);
            dfs(n,b,1);
            for(int i=0;i<=9;i++)
            {
                if(i)printf(" ");
                printf("%d",a[i]-b[i]);
            }
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7235963.html
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