题意：求一个区间内，满足连续的奇数长度是偶数，连续的偶数长度是奇数的数的个数。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-9;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 20 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int digit[MAXN];
LL dp[MAXN][2][MAXN];
LL dfs(int pos, int pre, int len, bool leadingzero, bool limit){//pos--当前位，pre--更高一位的数是奇还是偶，len--连续长度，leadingzero--是否有前导零，limit--当前位的数字是否有限制
if(!pos) return (pre & 1) != (len & 1);
if(!limit && dp[pos][pre][len] != -1) return dp[pos][pre][len];
LL ans = 0;
int up = limit ? digit[pos] : 9;
for(int i = 0; i <= up; ++i){
if(leadingzero){
if(i == 0){
ans += dfs(pos - 1, 0, 0, true, limit && i == up);
}
else{
ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);
}
}
else{
if(i & 1){
if(pre & 1){
ans += dfs(pos - 1, i & 1, len + 1, false, limit && i == up);
}
else{
if(len & 1){//若当前位是奇数，前一位是偶数，且已经有奇数长度的偶数，则可以继续延伸
ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);
}
}
}
else{
if(pre & 1){
if(!(len & 1)){
ans += dfs(pos - 1, i & 1, 1, false, limit && i == up);
}
}
else{
ans += dfs(pos - 1, i & 1, len + 1, false, limit && i == up);
}
}
}
}
if(!limit) dp[pos][pre][len] = ans;
return ans;
}
LL solve(LL x){
int cnt = 0;
while(x){
digit[++cnt] = x % 10;
x /= 10;
}
return dfs(cnt, 0, 0, true, true);
}
int main(){
int T;
scanf("%d", &T);
int kase = 0;
memset(dp, -1, sizeof dp);
while(T--){
LL L, R;
scanf("%lld%lld", &L, &R);
printf("Case #%d: %lld
", ++kase, solve(R) - solve(L - 1));
}
return 0;
}