版权声明：抱最大的希望，为最大的努力，做最坏的打算。 https://blog.csdn.net/qq_37748451/article/details/86588741
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题意：给出区间[m,n]，求区间内的所有数共有多少个0。
思路：设dp[i][j]表示处理到第i位时，它前面共有j个0（除了前导零）。
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[50][50];
int digit[51];
int cal(ll val) {
int len=0;
if(val<0)return 0;
if(val==0)return 1;
while(val) {
digit[len++]=val%10;
val/=10;
}
memset(dp,-1,sizeof(dp));
return len-1;
}
// 数位位置 状态 限制 前导0标志(false==是前导0)
ll dfs(int pos,int sta,int lim,bool J) {
if(pos<0)return (!J)?1:sta;
if((!lim) && (J) && (dp[pos][sta]^-1)) {//无限制且非前导0
return dp[pos][sta];
}
int len=lim?digit[pos]:9;
ll ans=0;
for(int i=0; i<=len; ++i) {
ans+=dfs(pos-1,sta+((J)&&(i==0)),lim&&(i==len),J||i);
}
if(lim==0&&(J))dp[pos][sta]=ans;//必须是非前导0
return ans;
}
ll work(ll val) {
if(val<0)return 0;
if(val==0)return 1;
return dfs(cal(val),0,1,0);
}
ll n,m;
int main() {
int T;
int coun=1;
scanf("%d",&T);
while(T--) {
scanf("%lld%lld",&n,&m);
printf("Case %d: %lld\n",coun++,work(m)-work(n-1));
}
return 0;
}