T1
懒的写。
T2
待补。
T3
待补。
T4
待补。
T5
简单数位 DP。不会数位 DP 建议先写 windy 数。
容易发现这个东西没有什么坑的点，直接写就行了。
#include <bits/stdc++.h>
#define llong long long
using namespace std;#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){x = 0; int w = 1;char ch = gc();while(ch < '0' || ch > '9'){if(ch == '-') w = -w;ch = gc();}while(ch >= '0' && ch <= '9')x = (x<<3)+(x<<1)+(ch^48), ch = gc();x *= w;
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){return read(x), read(y...);
}llong dp[18][10], Pow10[18];inline void prework(){Pow10[0] = 1;for(int i = 1; i <= 17; ++i) Pow10[i] = Pow10[i-1]*10;for(int i = 0; i <= 9; ++i) dp[0][i] = (i!=4);for(int i = 1; i <= 16; ++i){for(int j1 = 0; j1 <= 9; ++j1){if(j1 == 4) continue;for(int j2 = 0; j2 <= 9; ++j2){if(j1 == 1 && j2 == 3) continue;dp[i][j1] += dp[i-1][j2];}}}return;
}inline llong ctol(llong x){++x; llong res = 0;int lst = 0;for(int i = 16; i >= 1; --i){for(int j = 0; j <= 9; ++j){if(lst == 1 && j == 3) continue;if(dp[i][j] < x) x -= dp[i][j];else{res += j*Pow10[i], lst = j;break;}}}for(int i = 0; x; ++i)if((lst!=1 || i!=3) && i != 4)res = res-(res%10)+i, --x;return res;
}inline llong ltoc(llong x){llong res = 0;for(int i = 16; ~i; --i)if(x/Pow10[i]%10 == 4) return -1;for(int i = 15; ~i; --i)if(x/Pow10[i]%100 == 13) return -1;for(int i = 16; ~i; --i)for(int j = 0; j < x/Pow10[i]%10; ++j)if(x/Pow10[i+1]%10 != 1 || j != 3) res += dp[i][j];return res;
}int T;
int main(){// freopen("in.txt", "r", stdin);read(T);prework();while(T--){int op; llong x;read(op, x);if(op == 1) printf("%lld\n", ltoc(x));if(op == 2) printf("%lld\n", ctol(x));}return 0;
}