LightOJ 1140 How Many Zeroes？ [数位DP]【动态规划】

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1140

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D - How Many Zeroes?

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Description

Jimmy writes down the decimal representations of all natural numbers between and including m and n, (m ≤ n). How many zeroes will 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 and n, (m ≤ n).

Output

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

Sample Input

5

10 11

100 200

0 500

1234567890 2345678901

0 4294967295

Sample Output

Case 1: 1

Case 2: 22

Case 3: 92

Case 4: 987654304

Case 5: 3825876150

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题目大意 ：就是求区间[m,n]之间的所有数中 0的个数是多少 ?

解题思路 ： 、

标准的数位DP

dp[位数][0~9][0的个数];

然后记忆化搜索的时候

dfs(第几位,前面有没有非0的数,取数的限制,0的个数);

大概这样就能AC了 然后注意下细节就好了..

附本题代码

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#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MOD  = 1000000007;
const int maxn = 200010;

int num[30];
LL dp[30][12][20];

LL dfs(int pos,int pre,int limit,int m)
{
if(pos < 0) return m+(pre==0);

if(dp[pos][pre][m]!=-1&&!limit&&pre)
return dp[pos][pre][m];

int endi = 9;
if(limit) endi = num[pos];

LL res = 0;
for(int i=0;i<=endi;i++)
res+=dfs(pos-1,i||pre,limit&&(i==endi),m+(pre&&i==0));

if(!limit && pre) dp[pos][pre][m] = res;
return res;
}

LL solve(LL n)
{

int len = 0;

if(n==0) num[len++] = 0;

while(n)
{
num[len++] = n%10;
n/=10;
}

return dfs(len-1,0,1,0);
}

int main()
{
memset(dp,-1,sizeof(dp));
int _,p=0;
scanf("%d",&_);
while(_--)
{
LL n,m;
scanf("%lld%lld",&n,&m);
// printf("%lld %lld\n",solve(m),solve(n-1));
printf("Case %d: %lld\n",++p,solve(m)-solve(n-1));
}

return 0;
}