| Time Limit: 2000/1000 MS (Java/Others) | Memory Limit: 32768/32768 K (Java/Others) |
|---|---|
| Total Submission(s): 9715 | Accepted Submission(s): 5821 |
Problem Description8
A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string “13” and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.
Input
Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).
Output
Print each answer in a single line.
Sample Input
13
100
200
1000
Sample Output
1
1
2
2
Author
wqb0039
Source
2010 Asia Regional Chengdu Site —— Online Contest
第二道数位DP的题，题意很简单，输入一个数n，要求x ∈ [ 0, n ]，满足：x中含有能整除13并且x中含有"13"，输出满足条件的x的数量。就是这里的约束条件有两个，所以我们用于记忆化搜索的dp数组要设置为三维的。dp[ i ][ j ][ k ] 表示当前数位为i，j为当前位之前的各位和对13取模的值，k表示x中含有"1"，或者"13"或者都没有。当当前位为-1，也就是所有位都枚举完了，如果枚举出来的数满足条件的话，就可以计数1，否则计数0。
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 20;
int n;
int dp[maxn][13][3]; //前位和对13取模是多少 上一位是否为1 dp用来记录之前的状态
int num[maxn]; //记录各位数字
int dfs(int pos,int modsum,int isone,bool limit){
if(pos==-1) return (modsum==0)&&(isone==2); //如果满足条件就计数
if(!limit&&dp[pos][modsum][isone]!=-1) return dp[pos][modsum][isone];
int ans = 0; int up = limit ? num[pos] : 9; //判断是否前面都取最大值
for(int i=0;i<=up;i++){
int tmpmod = (modsum*10+i)%13; //加上当前之后的余数
int tmpisone = isone; //如果已经存在"13"就可以直接继续搜索了
if(isone==0&&i==1) tmpisone = 1;
else if(isone==1&&i==3) tmpisone = 2;
else if(isone==1&&i!=1) tmpisone = 0;
ans+=dfs(pos-1,tmpmod,tmpisone,limit&&(i==num[pos]));
}
if(!limit) dp[pos][modsum][isone] = ans;
return ans;
}
int solve(int x){
int pos = 0;
while(x){
num[pos++] = x%10;
x/=10;
}
int res = dfs(pos-1,0,0,true);
return res;
}
int main(){
memset(dp,255,sizeof(dp));
while(scanf("%d",&n)!=EOF){
printf("%d\n",solve(n));
}
return 0;
}