比赛链接

A - wwwvvvvvv

直接模拟

#include<cstdio>
#include<cstring>

const int N = 105;
int n, ans;
char s[N];

int main() {
	scanf("%s", s + 1);
	for(int i = 1; i <= strlen(s +1); i ++) {
		if(s[i] == 'v') ans ++;
		else ans += 2;
	}
	printf("%d", ans);
	return 0;
}

B - LOOKUP

枚举模拟。

#include<cstdio>
#include<cstring>

const int N = 105;
int n, m, ans;
char s[N], t[N];

int main() {
	scanf("%s %s", s + 1, t + 1);
	n = strlen(s + 1), m = strlen(t + 1);
	for(int i = 1, j; i  + m - 1 <= n; i ++) {
		for(j = 1; j <= m; j ++) {
			if(s[i + j - 1] != t[j]) break;
		}
		//printf("%d %d\n", i, j);
		if(j == m + 1) {
			puts("Yes");
			return 0;
		}
	}
	puts("No");
	return 0;
}

C - RANDOM

只要将字符数组每列都存储一下，再通过枚举其中一个字符数组的列，寻找在另一个字符数组中是否有与其相同的列即可。

#include<bits/stdc++.h>
using namespace std;
int main() {
	int h,w;
	cin >> h >> w;
	vector <string> s(w);
	vector <string> t(w);
	for(int i = 0; i < h; i ++) {
	    for(int j = 0; j < w; j ++) {
	    	s[j].resize(h), cin >> s[j][i];
		}
	}
	for(int i = 0; i < h; i ++) {
	    for(int j = 0; j < w; j ++) {
	    	t[j].resize(h), cin >> t[j][i];
		}
	}
	sort(s.begin(), s.end()), sort(t.begin(), t.end());
	for(int i = 0; i < w; i ++)  {
		if(s[i] != t[i]){
			puts("No");
    		return 0;
		}
	}
	puts("Yes");
    return 0;
}

D - Freefall

三分。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define int long long
long double a, b, c, d;
long double num(int x) {
	return b * x + a / sqrt(x + 1);
}
signed main() {
	scanf("%llf %llf", &a, &b), c = a;
	int l, r, mid;
	
	if(a < b) {
		mid = 1;
		while(1) {
			d = num(mid);
			if(d > c) {
				printf("%.10llf", c);
				return 0;
			}
			c = d, mid ++;
		}
	}
	else {
		l = 0, r = 1e18;
		while(l <= r) {
			mid = (r - l) / 2 + l;
			if(num(mid) < num(mid - 1) && num(mid) < num(mid + 1)) {
				printf("%.10llf", num(mid));
				return 0;
			}
			if(num(mid) < num(mid - 1)) l = mid + 1;
			else r = mid - 1;
		}
	}
	return 0;
}

E - Cheating Amidakuji

。。。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 2e5 + 5;
int n, m, a[N], b[N], c[N], ans[N];

int main() {
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; i ++) scanf("%d", &a[i]), c[i] = i;
	b[1] = 1;
	for(int i = 1, j; i < m; i ++) {
		j = b[i];
		if(j == a[i]) b[i + 1] = j + 1;
		else if(j == a[i] + 1) b[i + 1] = j - 1;
		else b[i + 1] = j;
	}
	for(int i = m; i; i --) {
		ans[i] = c[b[i]];
		swap(c[a[i]], c[a[i] + 1]);
	}
	for(int i = 1; i <= m; i ++) printf("%d\n", ans[i]);
	return 0;
}