D. Largest Digit
time limit per test: 1 second
memory limit per test: 1024 megabytes
Let f(x) be the largest digit in the decimal representation of a positive integer x. For example, f(4523)=5 and f(1001)=1.
Given four positive integers la, ra, lb and rb such that la≤ra and lb≤rb, calculate the maximum value of f(a+b), where la≤a≤ra and lb≤b≤rb.
Input
There are multiple test cases. The first line of the input contains an integer T (1≤T≤) indicating the number of test cases. For each test case:
The first and only line contains four integers la, ra, lb and rb (1≤la≤ra≤, 1≤lb≤rb≤).
Output
For each test case output one line containing one integer indicating the maximum value of f(a+b).
Example
input
2
178 182 83 85
2 5 3 6
output
7
9
Note
For the first sample test case, the answer is f(182+85)=f(267)=7.
For the second sample test case, the answer is f(4+5)=f(9)=9.
【思路分析】
签到题。显然,所求区间转换为[lb+la,rb+ra],遍历该区间会tle,当答案为9时进行剪枝即可。
cpp
#include <iostream>
#include <vector>
#include <unordered_map>
#include <map>
#include <cmath>
#include <algorithm>
#include <climits>
#include <stack>
#include <cstring>
#include <iomanip>
#include <set>
#include <queue>
#define i64 long long
using namespace std;
typedef pair<i64,i64> pii;
void solve() {
i64 la,ra,lb,rb,res = 0;
cin>>la>>ra>>lb>>rb;
for (i64 i = la+lb; i <= ra+rb; ++i) {
for (const auto &item: to_string(i)){
res = max(res,(i64)(item-'0'));
if (res == 9) goto flag;
}
}
flag:
cout<<res<<endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}