不要62Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 56596 Accepted Submission(s): 21994 Problem Description 杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer)。 Input 输入的都是整数对n、m(0<n≤m<1000000),如果遇到都是0的整数对,则输入结束。 Output 对于每个整数对,输出一个不含有不吉利数字的统计个数,该数值占一行位置。 Sample Input 1 100 0 0 Sample Output 80 Author qianneng Source Recommend lcy | We have carefully selected several similar problems for you: 2091 2093 2085 2082 2080 |
题解:数位DP
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<set>
#include<bitset>
using namespace std;
typedef unsigned long long ll;
ll dp[21][3]={0};
int a[25],len=0,n,m;
void fun(){
dp[0][0]=1;
for(int i=1;i<=20;i++){
dp[i][0]=dp[i-1][0]*9-dp[i-1][1];
dp[i][1]=dp[i-1][0];
dp[i][2]=dp[i-1][2]*10+dp[i-1][0]+dp[i-1][1];
}
}
ll gun(int n){
fun();
memset(a,0,sizeof(a));
len=1;
int nn=n;
while(n){
a[len++]=n%10;
n/=10;
}
int leap=0,last=0;
ll ans=0;
for(int i=len-1;i>=1;i--){
ans+=dp[i-1][2]*a[i];
if(leap)
ans+=dp[i-1][0]*a[i];
if(!leap&&a[i]>4)
ans+=dp[i-1][0];
if(!leap&&a[i+1]==6&&a[i]>2)
ans+=dp[i][1];
if(!leap&&a[i]>6)
ans+=dp[i-1][1];
if(a[i]==4||(a[i+1]==6&&a[i]==2))
leap=1;
}
return nn-ans;
}
int main(){
while(scanf("%d %d",&n,&m)!=EOF){
if(n==0&&m==0)
break;
ll ans1=gun(n);
ll ans2=gun(m+1);
printf("%llu\n",ans2-ans1);
}
return 0;
}