odd-even number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 661 Accepted Submission(s): 362
Problem Description
For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).
Input
First line a t,then t cases.every line contains two integers L and R.
Output
Print the output for each case on one line in the format as shown below.
Sample Input
2 1 100 110 220
Sample Output
Case #1: 29 Case #2: 36
Source
解题思路:因为要求奇数连续出现偶数次,偶数连续出现奇数次,均是跟数位有关的操作,考虑使用数位dp思路。
采用和以前数位dp相同的思路,关于数位dp可以看http://blog.youkuaiyun.com/wchhlbt/article/details/52119930
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 20;
ll n;
ll dp[maxn][20][2], bit[maxn];
ll dfs(int pos,int pre,int limit,int ok)//pos数字位数,pre
{
if(pos < 1 && ok) return 1;//如果顺利搜索到了第零位,说明这个数满足条件,返回1
if(pos < 1 && !ok) return 0;
if(!limit && dp[pos][pre][ok] != -1) return dp[pos][pre][ok];//记忆化搜索的关键
ll ret = 0;
int len = limit?bit[pos]:9;//如果处于非上界状态下,数字可以枚举0-9,处于上界状态则需要考虑给出的上界在该位的数字
for(int i = 0; i <= len; i++)
{
if(pre==11) {
if(i==0)
ret += dfs(pos-1, 11, limit&&i==len, 1);
else
ret += dfs(pos-1, i, limit&&i==len, (i+1)%2);
continue;
}
if(pre%2 != i%2)
{
if(ok==0)
continue;
else
{
ret += dfs(pos-1, i, limit&&i==len, (i+1)%2);
}
}
else
{
ret += dfs(pos-1, i, limit&&i==len, !ok);
}
//ret += dfs(pos-1, i, limit&&i==len);
}
if(!limit) dp[pos][pre][ok] = ret;
return ret;
}
ll solve(ll n)//预处理函数,将所给定的数字,处理为每个位置上的数字(上界)
{
int len = 0;
while(n)
{
bit[++len] = n%10;
n /= 10;
}
return dfs(len, 11, 1, 1);
}
int main()
{
int t;
ll a,b;
cin>>t;
memset(dp, -1, sizeof(dp));
for(int i = 1; i<= t ; i++)
{
cin>>a>>b;
//cout << solve(b) << ' ' << solve(a-1) << endl;
printf("Case #%d: %I64d\n",i,solve(b)-solve(a-1));
}
return 0;
}