lightoj - 1140 How Many Zeroes? （数位dp）

1140 - How Many Zeroes?



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Time Limit: 2 second(s)
Memory Limit: 32 MB

Jimmy writes down the decimal representations of all naturalnumbers between and including
m and n, (m ≤ n). How many zeroeswill he write down?

Input

Input starts with an integer T (≤ 11000),denoting the number of test cases.

Each case contains two unsigned 32-bit integers m andn, (m ≤ n).

Output

For each case, print the case number and the number of zeroeswritten down by Jimmy.

Sample Input

Output for Sample Input

5

10 11

100 200

0 500

1234567890 2345678901

0 4294967295

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

思路：dp[cur][s]表示到当前为止，前面有s个零的有多少答案

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=40;
int dig[maxn];
LL N,M;
LL dp[maxn][maxn];
int getsum(int pos)
{
    int sum=0;
    while(pos>=0)
    {
        sum=sum*10+dig[pos];
        pos--;
    }
    return sum;
}
LL dfs(int cur,int s,int e,int z)
{
    if(cur<0)return z?1:s;
    if(!e&&!z&&dp[cur][s]!=-1)return dp[cur][s];
    int end=e?dig[cur]:9;
    LL ans=0;
    for(int i=0;i<=end;i++)
    {
        if(z&&!i)ans+=dfs(cur-1,0,e&&(i==end),1);
        else if(i==0)
        {
            ans+=dfs(cur-1,s+1,e&&i==end,0);
        }
        else
            ans+=dfs(cur-1,s,e&&i==end,0);
    }
    if(!e&&!z)dp[cur][s]=ans;
    return ans;
}
LL solve(LL n)
{
    if(n<0)return 0;
    else if(n==0)return 1;
    int len=0;
    memset(dp,-1,sizeof(dp));
    while(n)
    {
        dig[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,1,1);
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&M,&N);
        printf("Case %d: %lld\n",cas++,solve(N)-solve(M-1));
    }
    return 0;
}