标签：dp
7. 369 numbers
A number is said to be a 369 number if
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
3
121 4325
432 4356
4234 4325667
Sample Output
60
58
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;
}
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SPOJ NUMTSN NUMTSN - 369 Numbers（数位dp）
标签：dp
原文地址：http://blog.csdn.net/u014737310/article/details/47041889
