PAT 1049. Counting Ones (30)

1049. Counting Ones (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (<=2³⁰).

Output Specification:

For each test case, print the number of 1's in one line.

Sample Input:

12

Sample Output:

5

先放我的代码。思路是逐位的数1

#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>#include<map>#include<vector>using namespace std;char a[100];int n;int tmp=n;int p=0;long long sum=0;int num(int y) {   int ssmm=0;   int p=0;   int k=1;   if(y<0) return 1;   while(p<=y)   {       ssmm+=k*(a[p]-'0');       k*=10;       p++;   }    return ssmm+1; }    void compute(int x)    {        int now=n;        int k=1;        for(int i=0;i<x;i++)            k*=10;        for(int i=0;i<x+1;i++)          now/=10;        sum+=now*k;        if(a[x]>'1') sum+=k;        else if(a[x]=='1')          sum+=num(x-1);    }  int main(){    cin>>n;    tmp=n;    while(tmp>0)    {        int zz=tmp%10;        tmp/=10;        a[p++]='0'+zz;    }      for(int i=0;i<p;i++)        compute(i);        cout<<sum;      return 0;}

附上别人的。思路一模一样，描述上更清晰。

#include<iostream>   using namespace std;  int main(){  int N,sum,radix,index;   cin >> N;  sum = 0;  radix = 1;  do{    index = N/radix%10;    if (index == 0)sum += N / (radix * 10)*radix;    /*例如2032 当取到index=0时，此位为1时X1XX ，前面有01两种可能，后面有00~99radix=100种可能，    PS：第一次做没有考虑到这种情况，借位有机会为1的情况*/    else if (index==1)    sum += N / (radix * 10) *radix+N%radix + 1;    /*例如24132，当取到1时，前面00~23种可能的时候，后面可以对应00~99radix=100种可能，而当取到24时，后面只有00~32=33种可能    PS:第一次做没有考虑到这种情况是只是最后一个的可能比较少*/    else if (index > 1)sum += (N / (radix * 10) + 1)*radix;    /*例如4542，当取到5时，前面0~4五种可能的时候，后面对应00~99种可能*/  radix *= 10;   }   while (N / radix!=0);  cout << sum<<endl;  system("pause");  return 0;}