题意：求一个区间里，不含有4或62的数字个数。
分析：
1、dp[i][j]---截止到第i位，当前第i位数字为j时不含4或62的数字个数。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int dp[10][15];
int tmp[10];
void init(){
dp[0][0] = 1;
for(int i = 1; i <= 6; ++i){//最多六位数，从低位到高位
for(int j = 0; j < 10; ++j){
for(int k = 0; k < 10; ++k){
if(j != 4 && !(j == 6 && k == 2)) dp[i][j] += dp[i - 1][k];//在dp[i - 1][k]的基础上第i位为j
}
}
}
}
int solve(int x){//solve函数计算的是1~x-1是否是不幸数的情况，并没有考虑数字x。
int cnt = 0;
while(x){
tmp[++cnt] = x % 10;
x /= 10;
}
tmp[++cnt] = 0;//防越界
int ans = 0;
for(int i = cnt - 1; i >= 1; --i){
for(int j = 0; j < tmp[i]; ++j){//只统计了1~x-1的情况
if(j != 4 && !(tmp[i + 1] == 6 && j == 2)){
ans += dp[i][j];
}
}
if(tmp[i] == 4 || (tmp[i + 1] == 6 && tmp[i] == 2)) break;//一旦当前枚举到的数中含4或62，则后面无论跟什么数字都不符合要求，停止枚举
}
return ans;
}
int main(){
int n, m;
init();
while(scanf("%d%d", &n, &m) == 2){
if(!n && !m) return 0;
printf("%d
", solve(m + 1) - solve(n));
}
return 0;
}
2、dfs的写法：
数位dp关键在于记录重复状态以便于记忆化搜索。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int dp[10][2], digit[10];
int dfs(int len, bool state, bool limit){
if(!len) return 1;
if(!limit && dp[len][state] != -1) return dp[len][state];
int ans = 0, up = limit ? digit[len] : 9;
for(int i = 0; i <= up; ++i){
if(i == 4 || state && i == 2) continue;
ans += dfs(len - 1, i == 6, limit && i == up);
}
if(!limit) dp[len][state] = ans;
return ans;
}
int solve(int x){
int cnt = 0;
while(x){
digit[++cnt] = x % 10;
x /= 10;
}
return dfs(cnt, false, true);
}
int main(){
int n, m;
memset(dp, -1, sizeof dp);
while(scanf("%d%d", &n, &m) == 2){
if(!n && !m) return 0;
printf("%d
", solve(m) - solve(n - 1));
}
return 0;
}