Codeforces Round 880 (Div. 2)题解
C. k-th equality
Consider all equalities of form a+b=c, where a has A digits, b has B digits, and c has C digits. All the numbers are positive integers and are written without leading zeroes. Find the k-th lexicographically smallest equality when written as a string like above or determine that it does not exist.
For example, the first three equalities satisfying A=1, B=1, C=2 are
- 1+9=10
- 2+8=10
- 2+9=11
An equality s is lexicographically smaller than an equality t with the same lengths of the numbers if and only if the following holds:
- in the first position where s and t differ, the equality s has a smaller digit than the corresponding digit in t
Input
Each test contains multiple test cases. The first line of input contains a single integer t (1≤t≤103) — the number of test cases. The description of test cases follows.
The first line of each test case contains integers A, B, C, k (1≤A,B,C≤6, 1≤k≤1012).
Each input file has at most 55 test cases which do not satisfy A,B,C≤3.
Output
For each test case, if there are strictly less than k valid equalities, output −1.
Otherwise, output the k-th equality as a string of form a+b=c.
Example
input
7
1 1 1 9
2 2 3 1
2 2 1 1
1 5 6 42
1 6 6 10000000
5 5 6 3031568815
6 6 6 1000000000000
output
2 + 1 = 3 10 + 90 = 100 -1 9 + 99996 = 100005 -1 78506 + 28543 = 107049 -1
题意：
给四个数 a b c k，找一个等式A+B=C，满足A是a位数，B是b位数，C是c位数，求满足条件字典序树第k小的等式
思路：
枚举A，每次找到可以到达的c的最小最大值，ans统计当前可以组成的等式个数
如果ans大于等于k，则目标状态一定是当前的a，然后通过 ans 和 k 差值，通过c的最大值减去差值 计算出 c 即可
代码：
#include
#define endl '\n'
#define debug(x) cout<<#x<<'='< pii;
typedef long long ll;
const int INF=0x3f3f3f3f;
void solve(){
int mn[10]={0,1,10,100,1000,10000,100000,10000000};
int a,b,c,k;
cin>>a>>b>>c>>k;
int ans=0;
for(int i=mn[a];i<10*mn[a];i++){ //枚举a
int l=i+mn[b],r=i+10*mn[b]-1; //计算出可到达的c
l=max(l,mn[c]); //和c的范围取交集
r=min(r,mn[c]*10-1);
ans+=max(0ll,r-l+1);
if(ans>=k){
int tmp=ans-k; //计算比目标状态大多少，用r减去
cout<