SPOJ NUMTSN NUMTSN - 369 Numbers（数位dp）

最新推荐文章于 2020-01-10 19:37:03 发布

原创最新推荐文章于 2020-01-10 19:37:03 发布 · 1.5k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#dp

DP 动态规划专栏收录该内容

142 篇文章

订阅专栏

本文介绍了一种特殊类型的整数——369数，这类数中3、6、9的数量相等且至少有一个。文章详细阐述了如何通过数位DP算法计算在指定区间内369数的数量，并提供了完整的C++实现代码。

NUMTSN - 369 Numbers

no tags

7. 369 numbers

A number is said to be a 369 number if

The count of 3s is equal to count of 6s and the count of 6s is equal to count of 9s.
The count of 3s is at least 1.

For Example 12369, 383676989, 396 all are 369 numbers whereas 213, 342143, 111 are not.

Given A and B find how many 369 numbers are there in the interval [A, B]. Print the answer modulo 1000000007.

Input

The first line contains the number of test cases (T) followed by T lines each containing 2 integers A and B.

Output

For each test case output the number of 369 numbers between A and B inclusive.

Constraints

T<=100

1<=A<=B<=10^50

Sample Input

121 4325

432 4356

4234 4325667

Sample Output

207159

链接:http://www.spoj.com/problems/NUMTSN/en/

/*
题意:求一段区间3 6 9 数目相同（且最少为1）的数的数目,
思路:注意到数非常大，有10^50次幂，所以用字符串保存
     然后数位dp  dp[i][j][k][m]分别代表到第i位，3 6 9 的数目


*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8

typedef long long ll;

using namespace std;
#define N 55
#define mod  1000000007

ll dp[N][N][N][N]; //
int bit[N];

void change(char *a)  //处理对于字符串减一操作
{
    int i,j;
    int len=strlen(a);
    len--;
    while(len>=0)
    {
        a[len]--;
        if(a[len]>='0') break;
        a[len]='9';
        len--;
    }
    len=strlen(a);
    for(i=0;i<len-1;i++) if(a[i]>='1') break;
    int k=0;
    for(i;i<len;i++)
        a[k++]=a[i];
    a[k]='\0';
}

ll dfs(int pos,int le,int mid,int ri,bool bound)
{
    if(pos==0) return le==mid&&mid==ri&&le>=1 ? 1:0;
    if(!bound&&dp[pos][le][mid][ri]>=0) return dp[pos][le][mid][ri];
    int up=bound ? bit[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(i==3)
          ans=(ans+dfs(pos-1,le+1,mid,ri,bound&&i==up))%mod;
        else
          if(i==6)
           ans=(ans+dfs(pos-1,le,mid+1,ri,bound&&i==up))%mod;
        else
            if(i==9)
             ans=(ans+dfs(pos-1,le,mid,ri+1,bound&&i==up))%mod;
        else
             ans=(ans+dfs(pos-1,le,mid,ri,bound&&i==up))%mod;
    }
    if(!bound) dp[pos][le][mid][ri]=ans;
    return ans;
}


ll solve(char *c)
{
   int i,j;
   int len=strlen(c);
   for(i=1;i<=len;i++)
    bit[i]=c[len-i]-'0';

   return dfs(len,0,0,0,true)%mod;
}

int main()
{
    int i,j,t;
    char a[N],b[N];
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",a,b);
        change(a);
        printf("%lld\n",(solve(b)-solve(a)+mod+mod)%mod);//必须+mod % mod 因为可能是负数
    }
    return 0;
}