WA_Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 15; 5 const int mod = 1e9+7; 6 7 ll l,r; 8 int n; 9 int a[15]; 10 ll dp[15][15]; 11 12 ll dfs(int pos,int pre,int sum, bool lead, bool limit){ 13 if( !pos ) return sum==0; 14 if( !limit && !lead && dp[pos][pre]!=-1 ) return dp[pos][pre]; 15 16 int up=limit?a[pos]:9; 17 ll ans = 0; 18 for(int i=0;i<=up;i++){ 19 ans += dfs(pos-1,i,(sum+i)%n,lead&&!i, limit&&(i==up)); 20 } 21 return (!limit&&!lead)?dp[pos][pre]=ans:ans; 22 } 23 24 ll solve(ll x){ 25 a[0] = 0; 26 while( x ) a[++a[0]]=x%10, x/=10; 27 return dfs(a[0],0,0,1,1); 28 } 29 30 int main() 31 { 32 33 while( ~scanf("%lld%lld%d",&l,&r,&n)){ 34 memset(dp,-1,sizeof(dp)); 35 printf("%lld ",solve(r)-solve(l-1)); 36 } 37 return 0; 38 }
AC_Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 15; 5 const int mod = 1e9+7; 6 7 ll l,r; 8 int n; 9 int a[15]; 10 ll dp[15][110]; 11 12 ll dfs(int pos,int pre,int sum, bool lead, bool limit){ 13 if( !pos ) return sum==0; 14 if( !limit && !lead && dp[pos][sum]!=-1 ) return dp[pos][sum]; 15 16 int up=limit?a[pos]:9; 17 ll ans = 0; 18 for(int i=0;i<=up;i++){ 19 ans += dfs(pos-1,i,(sum+i)%n,lead&&!i, limit&&(i==up)); 20 } 21 return (!limit&&!lead)?dp[pos][sum]=ans:ans; 22 } 23 24 ll solve(ll x){ 25 a[0] = 0; 26 while( x ) a[++a[0]]=x%10, x/=10; 27 return dfs(a[0],0,0,1,1); 28 } 29 30 int main() 31 { 32 33 while( ~scanf("%lld%lld%d",&l,&r,&n)){ 34 memset(dp,-1,sizeof(dp)); 35 printf("%lld ",solve(r)-solve(l-1)); 36 } 37 return 0; 38 }
AC代码和WA代码的不同之处在于：
WA: dp[pos][pre]:枚举到第pos位,前一个数是pre的满足题意的个数
AC: dp[pos][sum]:枚举到第pos位,前几位的和是sum的满足题意的个数
写这篇博客主要是想告诉自己dp不是怎么设都可以的，题目要求啥才设啥
这道题目跟pre没有直接关系，而是跟sum有关系