Codeforces Round 863 (Div. 3)

Codeforces Round 863 (Div. 3)

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Codeforces Round 863 (Div. 3)

A题

遍历这个字符串,如果要插入的数第一次小于当前的数,就将数插入到这里,如果到最后都没有插入数,插入到最后

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cstring>
#include <unordered_set>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <sstream>
#include <queue>
#define int long long
#define yes cout<<"YES"<<'\n'
#define no 	cout<<"NO"<<'\n'

using namespace std;
const int N = 100008;

void solve() {
	int n, x;
	scanf("%lld %lld", &n, &x);
	string s;
	cin >> s;
	int flag = 0;
	for (int i = 0; i < (int)s.size(); i++) {
		if (x > (s[i] - '0') && flag == 0) {//如果从来没插入过,并且大于当前的数字
			cout << x << s[i];//打印x和s[i]
			flag++;
		} else {
			cout << s[i];//否则只打印当前的s[i]
		}
	}
	if (flag == 0) {//没有打印过x
		cout << x;//在最后打印

	}
	cout << '\n';



}
signed main () {
	int t;
	cin >> t;
	while (t) {
		solve();
		t--;
	}


	return 0;
}

B题

比赛的时候看错题了,现在感觉好像看对也写不出来

这个题其实就是问两个点相距多少层,将两层相减取绝对值就可以.现在是怎么求出这个点所在的层数,可以发现点(x,y)和边界的距离最短的那个就是层数,就是min(x,y,n-x+1,n-y+1)的值.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cstring>
#include <unordered_set>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <sstream>
#include <queue>
#define int long long
#define yes cout<<"YES"<<'\n'
#define no 	cout<<"NO"<<'\n'

using namespace std;
const int N = 100008;

void solve() {
	int n;
	cin >> n;
	int x1, y1, x2, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	int a = min(min(x1, y1), min(n - x1 + 1, n - y1 + 1));//求x1,y1的层数
	int b = min(min(x2, y2), min(n - x2 + 1, n - y2 + 1));//求x2,y2的层数

	printf("%lld\n", abs(a - b));//打印abs(a-b);

}
signed main () {
	int t;
	cin >> t;
	while (t) {
		solve();
		t--;
	}


	return 0;
}