poj 3286 How many 0's?

How many 0's?

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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Description

A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?

Input

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.

Output

For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.

Sample Input

10 11100 2000 5001234567890 23456789010 4294967295-1 -1

Sample Output

122929876543043825876150

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呵呵基本和上一个题一样。。不过只求m——n之间所有数的0的个数

简单。。

排列组合，区间计数，计数序列,二分法,稳定婚姻问题,母函数

Source Code

Problem: E User: sdau_09_zysMemory: 164 KB Time: 110 MSLanguage: C++ Result: AcceptedPublic:No Yes

#include <iostream>#include <string>using namespace std;__int64 a,b;__int64 ansa[10],ansb[10];void count_digits(__int64 s,__int64 ans[],__int64 times=1)//求1-s之间的所有数中 0的个数，1的个数，2.....其中ans[0...9]为返回值{    __int64 i,d,p;    if (s <= 0)return ;    d = s % 10;    p = s / 10;    for (i = 1;i <= d;i ++)ans[i] += times;    while(p > 0)    {        ans[p % 10] += (d + 1) * times;        p = p / 10;    }    for (i = 0;i <= 9;i ++)ans[i] += times * (s / 10);    times *= 10;    count_digits((s / 10)-1,ans,times);    return ;}int main(){    //int i,j;    bool is;while(scanf("%I64d%I64d",&a,&b) != EOF){   if(a==-1&&b==-1)break;   if(!a&&!b){printf("1\n");continue;}   is=false;   memset(ansb,0,sizeof(ansb));   memset(ansa,0,sizeof(ansa));   if (a > b)   {      swap(a,b);   }   //if(a==b)is=true;   if(a==0)is=true;   if(a>0)a --;   if (b > a)   {      count_digits(b,ansb);      count_digits(a,ansa);   }   if(is)ansb[0]+=1;   //if(is)printf("%I64d\n",2*(ansb[0]-ansa[0]));else    printf("%I64d\n",ansb[0]-ansa[0]);}return 0;}