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不要62Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 58628 Accepted Submission(s): 23070 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: 2094 2090 2091 2093 2092 |
#include<bits/stdc++.h>
using namespace std;
int a[20], dp[20][2];
//dp[i][j]表示在第i个数位第j个状态时,有多少个满足题意的情况
int dfs(int pos, int pre, int state, bool limit){
if(pos == -1) return 1;//枚举结束
if(!limit && dp[pos][state] != -1) return dp[pos][state];
int up = limit ? a[pos] : 9;
int ans = 0;
for(int i=0; i<=up; i++){
if(i == 4) continue;
if(pre == 6 && i == 2) continue;
ans += dfs(pos-1, i, i == 6 ? 1 : 0, limit && a[pos] == i);
}
if(!limit) dp[pos][state] = ans;
return ans;
}
int solve(int x){//把数字的各个数位分解
int pos = 0;
while(x){
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos-1, 0, 0, true);
}
int main(){
int l, r;
while(scanf("%d%d", &l, &r) && (l || r)){
memset(dp, -1, sizeof(dp));
printf("%d\n", solve(r) - solve(l - 1));
}
return 0;
}