The Counting Problem - POJ 2282 数位dp
来源:互联网 发布:js中给文本框赋值 编辑:程序博客网 时间:2024/06/05 06:12
The Counting Problem
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3450 Accepted: 1838
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.
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 1044 497346 5421199 17481496 14031004 5031714 1901317 8541976 4941001 19600 0
Sample Output
1 2 1 1 1 1 1 1 1 185 185 185 185 190 96 96 96 95 9340 40 40 93 136 82 40 40 40 40115 666 215 215 214 205 205 154 105 10616 113 19 20 114 20 20 19 19 16107 105 100 101 101 197 200 200 200 200413 1133 503 503 503 502 502 417 402 412196 512 186 104 87 93 97 97 142 196398 1375 398 398 405 499 499 495 488 471294 1256 296 296 296 296 287 286 286 247
题意:求l到r的区间内所有数0-1的数字的个数。
思路:数位dp,具体实现看代码。
AC代码如下:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;//dp[i][j]表示i位数中j的个数,dp0[i]表示i位数中合法的0的个数ll dp[10][10],dp0[10],pow[15];int num[15],len,n;ll dfs(int pos,int N,int f)//f表示是否前面全是0,只对求0的个数有影响{ if(pos<=0) return 0; ll ans=0; int i,j,k,m=num[pos]; if(f==0) { ans+=m*dp[pos-1][N]; if(N<m) ans+=pow[pos-1]; else if(N==m) ans+=n%pow[pos-1]+1; ans+=dfs(pos-1,N,0); } else { ans+=(m-1)*dp[pos-1][N]; ans+=dfs(pos-1,N,0); ans+=dp0[pos-1]; } return ans;}ll solve(int p,int N){ if(p<=0) return 0; int i,j,k; len=0;n=p; while(p) { num[++len]=p%10; p/=10; } if(N==0) return dfs(len,N,1); else return dfs(len,N,0);}int main(){ int i,j,k,l,r; pow[0]=1; for(i=1;i<=9;i++) pow[i]=pow[i-1]*10; for(i=1;i<=9;i++) for(j=0;j<=9;j++) dp[i][j]=dp[i-1][j]*10+pow[i-1]; for(i=1;i<=9;i++) dp0[i]=9*dp[i-1][0]+dp0[i-1]; while(~scanf("%d%d",&l,&r) && l+r) { if(l>r) swap(l,r); printf("%I64d",solve(r,0)-solve(l-1,0)); for(i=1;i<=9;i++) printf(" %I64d",solve(r,i)-solve(l-1,i)); printf("\n"); }}
0 0
- The Counting Problem - POJ 2282 数位dp
- POJ 2282 The Counting Problem (数位dp)
- poj 2282 The Counting Problem (数位DP)
- poj 2282 The Counting Problem && poj 3286 How many 0's? (数位dp)
- 小白算法练习 hdu 不要62 POJ 2282 the Counting problem 数位dp
- POJ2282:The Counting Problem(数位DP)
- poj2282 The Counting Problem 数位dp
- poj 2282 The Counting Problem
- POJ-2282-The Counting Problem
- [POJ] 2282 -> The Counting Problem
- POJ 2282 The Counting Problem
- POJ 2282 The Counting Problem
- POJ 2282 The Counting Problem,组合数学
- UVa1640 - The Counting Problem(数位统计)
- uva 1640 The Counting Problem (数位dp||统计0-9的个数)
- POJ题目2282 The Counting Problem(数学)
- poj 2282 The Counting Problem 按位统计
- POJ 2282-The Counting Problem(组合数学_区间计数)
- fedora20 MP3解码器
- 利用ant脚本 自动构建svn全量/增量/减量 项目升级包
- hdu 4506 小明系列故事——师兄帮帮忙(快速幂入门)
- python第二部分:python脚本生成静态页面
- 浙大PAT考试1077~1080(2014上机复试题目)
- The Counting Problem - POJ 2282 数位dp
- LeetCode之旅(27)
- centos 修改host
- soj1001. Alphacode简单动规
- java反射类main方法
- Linux挂载windows的共享文件夹
- 阿里巴巴2014研发实习生笔试解析
- 深度学习深度信念网络DBNs—简易详解
- C++构建队列