题目地址： https://www.bnuoj.com/v3/problem_show.php?pid=52325
两份代码，解释在第二份代码里面
第一份代码整理一下看着爽
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 long long dp[20][11][3]; 6 int digit[20]; 7 int dfs(int len , int fore , int state , bool limit) 8 { 9 if (len == 0) 10 return 1; 11 if ( !limit && dp[len][fore][state] != -1 ) 12 return dp[len][fore][state]; 13 int res = 0; 14 int limitD = limit ? digit[len] : 9 ; 15 for (int j = 0 ; j<=limitD; j++) 16 { 17 int i; 18 if (fore == 10 && j == 0) i = 10; 19 else i = j; 20 if ( state == 0 ){ 21 if ( fore == 10 || i == fore ) 22 res += dfs( len - 1 , i , 0 , limit && i == limitD ); 23 } 24 else if ( state == 1 ){ 25 if ( fore == 10 || i <= fore ) 26 res += dfs( len - 1 , i , 1 , limit && i == limitD ); 27 } 28 else{ 29 if ( fore == 10 || i >= fore ) 30 res += dfs( len - 1 , i , 2 , limit && i == limitD ); 31 } 32 } 33 if (!limit) 34 dp[len][fore][state] = res; 35 return res; 36 } 37 long long solve(long long x) 38 { 39 int cnt = 0; 40 while (x) 41 { 42 digit[ ++cnt ] = ( int )( x % 10LL ); 43 x /= 10LL; 44 } 45 return dfs( cnt , 10 , 1 , true ) + dfs( cnt , 10 , 2 , true ) - dfs( cnt , 10 , 0 , true ); 46 } 47 int main() 48 { 49 memset(dp,-1,sizeof dp); 50 int N; 51 scanf("%d", &N ); 52 long long a,b; 53 while (N--) 54 { 55 scanf("%lld %lld" , &a , &b ); 56 printf("%lld ", solve( b ) - solve ( a - 1LL ) ); 57 } 58 }
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 //dp[i][j][k] 长度为i,前一位j,状态为k 6 //k=0 相等 k=1 递增 k=2 递减 7 long long dp[20][11][3]; 8 int digit[20]; 9 //dfs计算在digit限制下，长度为len，前一位为fore，的state状态的数字，特别的令fore==10代表前面全是0 10 int dfs(int len , int fore , int state , bool limit) 11 { 12 //最边界计数，即枚举到最后一位，0即为全是前导0。 13 if (len == 0) 14 { 15 return 1; 16 } 17 //记忆化搜索 18 if (!limit && dp[len][fore][state] != -1) 19 return dp[len][fore][state]; 20 int res = 0; 21 int limitD = limit?digit[len]:9; 22 for (int j = 0 ; j<=limitD; j++) 23 { 24 int i; 25 //确定填写的0,前导0变10,真0还是0 26 if (fore == 10 && j == 0) i = 10; 27 else i = j; 28 //如果前导0或者满足要求，那么继续dfs 29 //其中有些判断可以省略，但是所有的都加上fore==10,便于理解 30 if (state==0) //相等 31 { 32 if (fore==10||i==fore) 33 res += dfs(len-1,i,0,limit&&i==limitD); 34 } 35 else if (state==1)//递减 36 { 37 if (fore==10||i<=fore) 38 res += dfs(len-1,i,1,limit&&i==limitD); 39 } 40 else //递增 41 { 42 if (fore==10||i>=fore) 43 res += dfs(len-1,i,2,limit&&i==limitD); 44 } 45 } 46 if (!limit) dp[len][fore][state] = res; 47 return res; 48 } 49 long long solve(long long x) 50 { 51 int cnt = 0; 52 while (x) 53 { 54 digit[++cnt] = (int)(x % 10LL); 55 x /= 10LL; 56 } 57 //递增+递减-相等 58 return dfs(cnt,10,1,true) + dfs(cnt,10,2,true) - dfs(cnt,10,0,true); 59 } 60 int main() 61 { 62 memset(dp,-1,sizeof dp); 63 int N; 64 scanf("%d",&N); 65 long long a,b; 66 while (N--) 67 { 68 scanf("%lld%lld",&a,&b); 69 printf("%lld ",solve(b) - solve(a-1LL) ); 70 } 71 }