R2 day2

简单写一下吧

emmmm，来晚了1h，没赶上，所以没交.......(捂脸

T1

开始读错题了诶
开烤1.2h后
发现是个傻逼题....
排序一下，维护前缀最左,右端点
随机数据我跑的比他们都慢..........

#include<cstdio>
#include<algorithm> 
#define LL long long 
const int maxn = 2000007; 
struct node { 
	int v,loc; 
} a[maxn]; 
bool cmp(node q,node b) { 
	if(q.v == b.v) return q.loc <= b.loc; 
	return q.v > b.v; 
} 
int n,m;
int main() { 
	freopen("w.in","r",stdin); freopen("w.out","w",stdout); 
	LL ans = 0; scanf("%d",&n); 
	for(int i = 1;i <= n;++ i) scanf("%d",&a[i].v), a[i].loc = i; 
	std::sort(a + 1, a + n + 1,cmp); 
	int l = a[1].loc,r = a[1].loc; 
	ans = a[1].v; 
	for(int i = 2;i <= n;++ i) { 
		int k = a[i].loc; 
		if(k < l) l = k; 
		if(k > r) r = k; 
		if(k >= l)ans = std::max(1ll * (k - l + 1) * a[i].v,ans); 
		if(k <= r)ans = std::max(1ll * (r - k + 1) * a[i].v,ans); 
	} 
 	printf("%I64d
",ans); 
    return 0;
} 

T2

emmmm，推了一下式子
发现这个

所以只需要尽可能的合并就行了

#include <cstdio> 
int cnt[10000007]; 
const int mod = 998244353; 
#define LL long long 
int main() {
    freopen("s.in","r",stdin);
    freopen("s.out","w",stdout);
    int n;
    scanf("%d",&n);
    int cnt[32] = {0}; 
    for(int i = 1;i <= n;i ++) { 
        int x; scanf("%d", &x); 
        for(int j = 0;j < 32;j ++) if(x & (1 << j)) cnt[j] ++; 
    } 
    LL ans = 0; 
    for(int i = 1;i <= n;i++) { 
        int x = 0; 
        for (int j = 0;j < 32;j ++) if (cnt[j]) x |= (1 << j), cnt[j] --; 
        ans = (ans + 1ll * x * x) % mod; 
    }   
    printf("%lld
",ans % mod); 

    return 0;
}

T3

状压dp
记录本行和上一行状态
然后把可行状态抽出来就行了

#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
const int mod = 1e9 + 7; 
inline int read() { 
	int x = 0; 
	char c = getchar(); 
    while(c < '0' || c > '9')c = getchar(); 
    while(c <= '9'&& c >= '0')x = x * 10 + c - '0',c = getchar();return x;
	return x; 
} 
int n,m,k; 
const int maxn = 17; 
int cant[maxn]; 
bool judge(int s) { 
	if((s & (s >> 1)) || (s & (s >> 2))) return false; 
	return true; 
} 
int s[maxn]; 
int id[maxn]; 
int dp[maxn][407][407]; 
int main() { 
  	n = read() , m= read() - 1,k = read(); 
  	for(int t,p,i = 1;i <= k;++ i) { 
  		t = read(),p = read() - 1; 
  		cant[t] |= (1 << p); 
  	} 
  	int T = (1 << m) - 1; 
  	int cnt = 0; 
  	for(int i = 0;i <= T;++ i) 
  		if(judge(i)) s[++ cnt] = i,s[i] = cnt; 
	for(int i = 0;i <= T;++ i) dp[0][0][i] = 1; 
	long long ans = 0; 
	for(int i = 1;i <= n;++ i) { 
		for(int s1 = 0; s1 <= T;++ s1) { 
			if(cant[i] & s[s1]) continue; 
			for(int s2 = 0;s2 <= (i <= 1 ? 1 : T);++ s2) { 
				if(cant[i - 1] & s2 || ((s1 >> 1) & s2)  || ((s1 << 1) & s2) || (s1 & s2)) continue; 
				for(int s3 = 0;s3 <= (i <= 2 ? 1 : T);++ s3) { 
					if((s3 & s1) || (cant[i - 2] & s3) || ((s2 >> 1) & s3)  || ((s2 << 1) & s3) || (s2 & s3) ) continue; 
					dp[i][s2][s1] += dp[i - 1][s3][s2]; 
					dp[i][s2][s1] %= mod; 
				} 
				if(i == n) {ans += dp[i][s2][s1],ans %= mod;} 
			} 
		} 
	} 
	printf("%lld
",ans); 
	return 0; 
}