fzu 2113 数位dp

#include<stdio.h>
#include<string.h>
#define N 20
#define ll __int64
ll dp[N][N];//最多记忆400种情况
ll digit[N];
ll dfs(ll len,ll cnt,ll ok) {
 if(!len) return cnt;
 if(!ok&&dp[len][cnt]!=-1)
    return dp[len][cnt];
 ll i,ans=0,maxx=ok?digit[len]:9;
 for(i=0;i<=maxx;i++) {
    if(i==1)ans+=dfs(len-1,cnt+1,ok&&i==maxx);
    else
        ans+=dfs(len-1,cnt,ok&&i==maxx);
 }
 if(!ok)
    dp[len][cnt]=ans;
 return ans;
}
ll f(ll n) {
ll len=0;
while(n) {
    digit[++len]=n%10;
    n/=10;
}
return dfs(len,0,1);
}
int main() {
      ll n,m,k;
      memset(dp,-1,sizeof(dp));
      while(scanf("%I64d%I64d",&n,&m)!=EOF) {
            if(n>m) {
                k=n;
                n=m;
                m=k;
            }
        printf("%I64d\n",f(m)-f(n-1));
      }
return 0;}

<pre name="code" class="cpp">/*
数位dp
直接相加更快
另一种思路
*/
#include<stdio.h>
#include<string.h>
#define N 20
#define ll __int64
ll lower[N];//储存十的倍数。
ll dp[N][10];
ll digit[N];
ll  rest[N];//用来储存有多少数
ll dfs(ll len,ll cnt,ll ok) {
   if(!len) return cnt==1?1:0;
   if(!ok&&dp[len][cnt]!=-1)
    return dp[len][cnt];
   ll i,ans=0,maxx=ok?digit[len]:9;
   for(i=0;i<=maxx;i++)
    ans+=dfs(len-1,i,ok&&i==maxx);
   if(cnt==1) {
    if(!ok)
        ans+=lower[len];
    else
        ans+=rest[len];//
   }
   if(!ok)
    dp[len][cnt]=ans;
   return ans;
}
ll f(ll n) {
ll len=0;
ll z=n;
while(n) {
    digit[++len]=n%10;
    rest[len]=z%lower[len]+1;//比如是163的话，因为从100-163所以取余后要加一
    n/=10;
}
return dfs(len,0,1);
}
int main() {
      ll i,n,m;
      lower[0]=1;
      memset(dp,-1,sizeof(dp));
      for(i=1;i<=19;i++)
        lower[i]=lower[i-1]*10;
      while(scanf("%I64d%I64d",&n,&m)!=EOF) {
        printf("%I64d\n",f(m)-f(n-1));
      }
return 0;}