HZOI20190906模拟38 金，斯诺，赤

题面：https://www.cnblogs.com/Juve/articles/11479415.html

T1：高精度gcd，其实不用写高精度取模，gcd还有一种求法

int gcd(int a,int b){
	if(a==b) return a;
	if(a%2==0&&b%2==0) return 2*gcd(a/2,b/2);
	if(a%2==0) return gcd(a/2,b);
	if(b%2==0) return gcd(a,b/2);
	if(a<b) swap(a,b);
	return gcd(a-b,b);
}

然后愉快地AC

#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
#define re register
using namespace std;
int t,la,lb,c[105];
char a[105],b[105];
struct bigint{
	int m[105];
	bigint(){memset(m,0,sizeof(m));}
	inline friend void operator *= (bigint &a,re int b){
		int x=0;
		for(re int i=1;i<=a.m[0];i++){
			re int y=a.m[i]*b+x;
			a.m[i]=y%10;
			x=y/10;
		}
		while(x){
		    a.m[++a.m[0]]=x%10;
		    x/=10;
		}
	}
	inline friend void operator /= (bigint &a,re int b){
		re int x=0;
		for(re int i=a.m[0];i>=1;i--){
			x+=a.m[i];
			a.m[i]=x/b;
			x%=b;
			x*=10;
		}
		while(a.m[a.m[0]]==0&&a.m[0]>1)
			a.m[0]--;
	}
	inline friend bigint operator - (bigint a,bigint b){
		bigint c;
		re int i=1;
		while((i<=a.m[0])||(i<=b.m[0])){
			if(a.m[i]<b.m[i]){
				a.m[i]+=10;
				a.m[i+1]--;
			}
			c.m[i]=a.m[i]-b.m[i];
			i++;
		}
		while(c.m[i]==0&&i>1)
			i--;
		c.m[0]=i;
		return c;
	}
	inline friend bool operator >= (bigint a,bigint b){
		if(a.m[0]>b.m[0]) return 1;
		if(a.m[0]<b.m[0]) return 0;
		for(int i=a.m[0];i>=1;--i){
			if(a.m[i]==b.m[i]) continue;
			return a.m[i]>b.m[i];
		}
		return 1;
	}
	inline friend bool operator == (bigint a,bigint b){
		int p=a.m[0],q=b.m[0];
		if(p!=q) return 0;
		for(int i=1;i<=p;++i){
			if(a.m[i]!=b.m[i]) return 0;
		}
		return 1;
	}
	inline friend void print(bigint a){
		for(re int i=a.m[0];i>=1;i--)
			printf("%lld",a.m[i]);
		puts("");
	}
}n,m;
bool judge(bigint a){
	int p=a.m[1];
	//cout<<p<<endl;
	if(p%2==0) return 1;
	return 0;
}
bool check(bigint a,bigint b){
	//print(a),print(b);
	if(a==b){
		if(a.m[0]==1&&a.m[1]==1) return 1;
		else return 0;
	}
	bool p=judge(a),q=judge(b);
	//cout<<p<<' '<<q<<endl;
	if(p&&q) return 0;
	if(p){
		a/=2;
		return check(a,b);
	}
	if(q){
		b/=2;
		return check(a,b);
	}
	if(!(a>=b)) swap(a,b);
	return check(a-b,b);
}
int gcd(int a,int b){
	if(a==b) return a;
	if(a%2==0&&b%2==0) return 2*gcd(a/2,b/2);
	if(a%2==0) return gcd(a/2,b);
	if(b%2==0) return gcd(a,b/2);
	if(a<b) swap(a,b);
	return gcd(a-b,b);
}
signed main(){
	scanf("%lld",&t);
	while(t--){
		memset(n.m,0,sizeof(n.m));
		memset(m.m,0,sizeof(m.m));
		scanf("%s %s",a+1,b+1);
		la=strlen(a+1),lb=strlen(b+1);
		n.m[0]=la,m.m[0]=lb;
		for(int i=1;i<=la;++i) n.m[la-i+1]=a[i]-'0';
		for(int i=1;i<=lb;++i) m.m[lb-i+1]=b[i]-'0';
		if(check(n,m)==1) puts("Yes");
		else puts("No");
	}
	return 0;
}

T2：

正解$O(n)$，数状数组卡常可A

$sum_i$表示前缀和

我们对于每一个i，求出$sum_i-sum_{j-1}>\frac{i-j+1}{2}(j<i)$的数量

然后就转化成了$2sum_i-i>2sum_{j-1}-(j-1)$，然后数状数组即可

然后用总区间减去不合法的即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define re register
using namespace std;
const int MAXN=5e6+5;
int n,sum0[MAXN],sum1[MAXN],sum2[MAXN],sum[MAXN];
char a[MAXN],ch;
long long ans=0;
struct BIT_tree{
	int c[MAXN*3];
	inline int lowbit(re int x){
		return x&-x;
	}
	inline void add(re int pos){
		//cout<<pos<<' '<<val<<endl;
		while(pos<=2*n){
			++c[pos];
			pos+=lowbit(pos);
			//cout<<pos<<endl;
		}
	}
	inline int query(re int pos){
		re int res=0;
		while(pos>0){
			res+=c[pos];
			pos-=lowbit(pos);
		}
		return res;
	}
}tr[3];
signed main(){
	//freopen("ex4.in","r",stdin);
	scanf("%d",&n);
	scanf("%s",a+1);
	for(re int i=1;i<=n;++i){
		sum0[i]=sum0[i-1],sum1[i]=sum1[i-1],sum2[i]=sum2[i-1];
		if(a[i]=='0') ++sum0[i];
		if(a[i]=='1') ++sum1[i];
		if(a[i]=='2') ++sum2[i];
		tr[0].add(2*sum0[i-1]-i+n+2);
		ans+=tr[0].query(2*sum0[i]-i+n);
		tr[1].add(2*sum1[i-1]-i+n+2);
		ans+=tr[1].query(2*sum1[i]-i+n);
		tr[2].add(2*sum2[i-1]-i+n+2);
		ans+=tr[2].query(2*sum2[i]-i+n);
	}
	printf("%lld\n",(1ll*n*(n+1)/2)-ans);
	return 0;
}

%%正解大佬gby

#include <iostream>
#include <cstring>
#include <cstdio>
#define lowbit(x) ((x)&(-(x)))
#define N 5555555
#define LL long long
#define pre(i,j) pre[(i)+nn][j]
#define tb(i,j) tb[(i)+nn][j]

using namespace std;

int nn,pre[2*N][3],tb[2*N][3];
char arr[N];
int dat[N][3];
LL ans;
int main(){
	scanf("%d",&nn);
	scanf("%s",arr+1);
	for(int i=1;i<=nn;i++){
		dat[i][0]=dat[i-1][0]+1;
		dat[i][1]=dat[i-1][1]+1;
		dat[i][2]=dat[i-1][2]+1;
		dat[i][arr[i]-'0']-=2;
	}
	/*for(int i=0;i<=2;i++){
		printf("%d:",i);
		for(int j=1;j<=nn;j++)
			cout<<dat[j][i]<<" ";
		cout<<endl;
		}*/
	ans=(long long)nn*(nn+1)/2;
	//cout<<ans<<endl;
	pre(0,0)++,pre(0,1)++,pre(0,2)++;
	tb(0,0)++ ,tb(0,1)++, tb(0,2)++;
	static int lst[3];
	for(int i=1;i<=nn;i++){
		for(int k=0;k<=2;k++){
			tb(dat[i][k],k)++;
			if(dat[i][k]==lst[k]+1){
				pre(dat[i][k],k)=pre(lst[k],k)+tb(dat[i][k],k);
			}
			else {//dat[i][k]==lst[k]-1
				pre(dat[i][k],k)=pre(dat[i][k]-1,k)+tb(dat[i][k],k);
				pre(lst[k],k)=pre(dat[i][k],k)+tb(lst[k],k);
			}
			ans-=i+1-pre(dat[i][k],k);
			//	cout<<k<<":"<<i+1-pre(dat[i][k],k)<<endl;
		}
		lst[0]=dat[i][0];
		lst[1]=dat[i][1];
		lst[2]=dat[i][2];
		/*for(int i=0;i<=2;i++){
			printf("%d:",i);
			for(int j=-nn;j<=nn;j++)
				cout<<pre(j,i)<<" ";
			cout<<endl;
			}*/
	}
	cout<<ans<<endl;
}

T3:

wqs二分

首先一个暴力dp：

#include<iostream>
#include<cstdio>
#include<cstring>
#define re register
using namespace std;
const int MAXN=100005;
int n,a,b;
double p[MAXN],q[MAXN],ans=0.0,f[2][605][605];
inline double max(re double a,re double b){
	return a>b?a:b;
}
signed main(){
	while(~scanf("%d%d%d",&n,&a,&b)){
		//memset(f,0,sizeof(f));
		for(re int i=1;i<=n;++i) scanf("%lf",&p[i]);
		for(re int i=1;i<=n;++i) scanf("%lf",&q[i]);
		ans=0.0;
		//f[0][0][0]=0;
		memset(f[0],0,sizeof(f[0]));
		for(re int i=1;i<=n;++i){
			for(re int j=0;j<=min(a,n);++j){
				for(re int k=0;k<=min(b,n);++k){
				    f[i&1][j][k]=f[i&1^1][j][k];
				    if(j!=0) f[i&1][j][k]=max(f[i&1][j][k],f[i&1^1][j-1][k]+p[i]);
				    if(k!=0) f[i&1][j][k]=max(f[i&1][j][k],f[i&1^1][j][k-1]+q[i]);
				    if(j!=0&&k!=0)
				        f[i&1][j][k]=max(f[i&1][j][k],f[i&1^1][j-1][k-1]+p[i]+q[i]-p[i]*q[i]);
				}
			}
		}
		printf("%0.3lf\n",f[n&1][a][b]);
	}
	return 0;
}

然后优化：

#include<iostream>
#include<cstdio>
#include<cstring>
#define re register
#define eps 1e-8
using namespace std;
const int MAXN=100005;
int n,a,b;
double p[MAXN],q[MAXN],ans=0.0,l,r,L,R,f[MAXN],fa[MAXN],fb[MAXN];
inline double max(re double a,re double b){
	return a>b?a:b;
}
bool judge(double na,double nb){
	memset(f,0,sizeof(f));
	memset(fa,0,sizeof(fa));
	memset(fb,0,sizeof(fb));
	for(int i=1;i<=n;i++){
        f[i]=f[i-1],fa[i]=fa[i-1],fb[i]=fb[i-1];
        if(f[i-1]+p[i]>f[i]+na)
			f[i]=f[i-1]+p[i]-na,fa[i]=fa[i-1]+1,fb[i]=fb[i-1];
        if(f[i-1]+q[i]>f[i]+nb)
			f[i]=f[i-1]+q[i]-nb,fb[i]=fb[i-1]+1,fa[i]=fa[i-1];
        if(f[i-1]+p[i]+q[i]-p[i]*q[i]>f[i]+na+nb)
			f[i]=f[i-1]+p[i]+q[i]-p[i]*q[i]-na-nb,fa[i]=fa[i-1]+1,fb[i]=fb[i-1]+1;
    }
	return fb[n]>b;
}
bool check(double na){
	L=0.0,R=1.0;
	while(R-L>eps){
		double mid=(L+R)/2.0;
		if(judge(na,mid)) L=mid;
		else R=mid;
	}
	judge(na,R);
	return fa[n]>a;
}
signed main(){
	while(~scanf("%d%d%d",&n,&a,&b)){
		for(re int i=1;i<=n;++i) scanf("%lf",&p[i]);
		for(re int i=1;i<=n;++i) scanf("%lf",&q[i]);
		l=0.0,r=1.0;
		ans=0.0;
		memset(f,0,sizeof(f));
		memset(fa,0,sizeof(fa));
		memset(fb,0,sizeof(fb));
		while(r-l>eps){
			double mid=(l+r)/2.0;
			if(check(mid)) l=mid;
			else r=mid;
		}
		judge(r,R);
		printf("%0.5lf\n",f[n]+a*r+b*R);
	}
	return 0;
}