Codeforces Edu Round 63 A-E

A. Reverse a Substring

容易看出，只要符合递增顺序就符合\(NO\)，否则则可以找到一组，每次记录最大值比较即可。

#include <cstdio>
#include <iostream>
using namespace std;
const int N = 300010;
int n;
char s[N];
int main(){
	scanf("%d%s", &n, s + 1);
	int maxn = s[1], k = 1;
	for(int i = 2; i <= n; i++){
		if(s[i] < maxn){
			printf("YES\n%d %d\n", k, i);
			return 0;
		}
		if(s[i] > maxn){
			maxn = s[i];
			k = i;
		}
	}
	printf("NO\n");
	return 0;
} 

B. Game with Telephone Numbers

\(Vasya\)用尽所有轮次可以把几个\(8\)扔到最前面，只要这个数大于轮次，就说明\(Petya\)无法逆天改命。

#include <cstdio>
#include <iostream>
#include <cstring> 
using namespace std;
const int N	= 100010;
int n;
char s[N];
int main(){
	scanf("%d%s", &n, s + 1);
	int round = (n - 11) >> 1;
	int cnt = 0; 
	for(int i = 1; i <= round && round <= n; i++){
		if(s[i] == '8')cnt++, round ++;
	}
	int i = round + 1;
	while(s[i] == '8' && i <= n) i++, cnt++;
	if(cnt > (n - 11) >> 1) puts("YES");
	else puts("NO");
	return 0;
}

C. Alarm Clocks Everywhere

显然，y设置为\(x[1]\)即可，因为即使设置在之前，还是要经过\(x[1]\)，还是以\(p\)往上跳，是没有区别的。

我们只要找到一个\(p_j\)，使其满足\(p_j | a[i + 1] - a[i] (1 <= i < n)\)。（整除的意思）

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 300010;
int n, m;
LL x[N], p[N];
LL gcd(LL a, LL b){
	return b ? gcd(b, a % b) : a;
}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) 
		scanf("%lld", x + i);
	for(int i = 1; i <= m; i++)
		scanf("%lld", p + i);
		
	LL ans = x[2] - x[1];
	for(int i = 2; i < n; i++)
		ans = gcd(ans, x[i + 1] - x[i]);
	
	for(int i = 1; i <= m; i++){
		if(ans % p[i] == 0){
			printf("YES\n%lld %d\n", x[1], i);
			return 0;
		}
	}
	printf("NO");
	return 0;
}

D. Beautiful Array

想到了对数列总和最大贡献，但是其实最大贡献不一定能最大化答案，于是惨遭\(WA10\)。

\(WA\)炸代码：

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 300010, INF = 2147483647;
typedef long long LL;
int n, x, L[N];
LL a[N], f[N];
int main(){
	scanf("%d%d", &n, &x);
	for(int i = 1; i <= n; i++) 
		scanf("%lld", a + i);
	
	if(x < 0){
		f[1] = a[1];
		L[1] = 1;
		LL minn = f[1];
		int k = 1;
		for(int i = 2; i <= n; i++) {
			if(f[i - 1] + a[i] < a[i]){
				L[i] = L[i - 1];
				f[i] = f[i - 1] + a[i];
			}else{
				L[i] = i;
				f[i] = a[i];
			}
			
			if(f[i] < minn){
				minn = f[i], k = i;
			}
		}
		if(minn < 0){
			for(int i = L[k]; i <= k; i++) 
				a[i] *= x;
		} 
	}else{
		
		f[1] = a[1];
		L[1] = 1;
		LL maxn = f[1];
		int k = 1;
		for(int i = 2; i <= n; i++) {
			if(f[i - 1] + a[i] > a[i]){
				L[i] = L[i - 1];
				f[i] = f[i - 1] + a[i];
			}else{
				L[i] = i;
				f[i] = a[i];
			}
			
			if(f[i] > maxn){
				maxn = f[i], k = i;
			}
		}
		if(maxn > 0){
			for(int i = L[k]; i <= k; i++) 
				a[i] *= x;
		} 
		
	}
	
	f[1] = a[1];
	LL maxn = f[1];
	for(int i = 2; i <= n; i++){
		f[i] = max(f[i - 1] + a[i], a[i]);
		maxn = max(maxn, f[i]);
	}
	printf("%lld\n", max(maxn, 0ll));
	return 0;
}

看了题解直接自闭掉，实际上，可以设

\(f[0]\)为当前以\(i\)为右端点，未加入\(x\)的极大值，那么它必须大于\(0\)

\(f[1]\)代表开始连续计算\(* x\)了，如果这时候还没有\(f[0]\)大，那么它也没用了

\(f[2]\)代表乘完了的，探究是否还能再加入\(a[i]\)

如果\(x > 0\)，那么\(f[2] + a[i]\)一定不会最优

如果\(x < 0\)，有可能有这种情况，前面负数后面正数

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 300010, INF = 2147483647;
typedef long long LL;
int n, x, L[N];
LL a[N], f[3], ans = 0;
int main(){
	scanf("%d%d", &n, &x);
	for(int i = 1; i <= n; i++) 
		scanf("%lld", a + i);
	
	for(int i = 1; i <= n; i++){
		f[0] = max(f[0] + a[i], 0ll);
		f[1] = max(f[0], f[1] + a[i] * x);
		f[2] = max(f[1], f[2] + a[i]);
		ans = max(ans, f[2]); 
	} 
	printf("%lld\n", ans);
	return 0;
}

E. Guess the Root

高斯消元\(/\)拉格朗日插值法都不会，咕咕咕。