uva 1640 The Counting Problem （数位dp||统计0-9的个数）

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 rst

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

题意： 统计（n，m） 统计0-9的个数

注意前导零， 统计零时候需要特判一下

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int dp[11][11],pre[11];void work(int num){int i,j,k;memset(dp,0,sizeof(dp));for(i=1;i<=8;i++) {for(j=0;j<10;j++) {if(j==num) dp[i][j]=pow(10.0,i-1);for(k=0;k<10;k++) {dp[i][j]+=dp[i-1][k];}}}}int solve(int num,int num2){memset(pre,0,sizeof(pre));int ans,bit[10],k=0;while(num) {bit[++k]=num%10;num/=10;pre[k]+=pre[k-1]+bit[k]*pow(10.0,k-1);}ans=0;for(int i=k;i>0;i--) {for(int j=0;j<bit[i];j++) {    ans+=dp[i][j];}if(bit[i]==num2) ans+=pre[i-1]+1;}if(num2==0) {for(int i=k;i>0;i--) ans-=pow(10.0,i-1);}return ans;}int main(){int l,r;while(cin>>l>>r) {if(l>r) swap(l,r);if(l==0 && r==0) break;for(int i=0;i<10;i++) {work(i);if(i) cout<<" ";cout<<solve(r,i)-solve(l-1,i);}cout<<endl;}return 0;}