简单的数位DP。。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 500005
#define maxm 300005
#define eps 1e-3
#define mod 9999677
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head
LL dp[20][20];
int digit[maxn];
LL dfs(int pos, int left, int limit)
{
if(dp[pos][left] != -1&& !limit) return dp[pos][left];
if(pos == 0) {
if(left == 0) return 1;
else return 0;
}
int a = 0, b = limit ? digit[pos] : 9;
LL ans = 0;
for(int i = a; i <= b; i++) ans += dfs(pos-1, (left + i) % 10, limit & i == b);
if(!limit) dp[pos][left] = ans;
return ans;
}
void work()
{
LL l, r;
scanf("%lld%lld", &l, &r);
l--;
int cnt = 0;
while(r) digit[++cnt] = r % 10, r /= 10;
LL ans = dfs(cnt, 0, 1);
cnt = 0;
while(l) digit[++cnt] = l % 10, l /= 10;
ans -= dfs(cnt, 0, 1);
printf("%lld\n", ans);
}
int main()
{
int _, __;
memset(dp, -1, sizeof dp);
while(scanf("%d", &_)!=EOF) {
__ = 0;
while(_--) printf("Case #%d: ", ++__), work();
}
return 0;
}
