一套模拟赛

昨晚的模拟赛很好啊

T1:Super GCD

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
struct bignum{
    int a[205],len;
    bignum(){
        memset(a,0,sizeof a);
        len=0;
    }
    bignum(int x){
        memset(a,0,sizeof a);len=0;
        while(x){
            a[++len]=x%10;
            x/=10;
        }
    }
    void div(){
        for(int i=len;i;i--){
            if((i>1)&&(a[i]&1))a[i-1]+=10;
            a[i]>>=1;
        }
        while(len>1&&!a[len])len--;
    }
    bignum operator - (bignum b){
        bignum c;
        c.len=max(len,b.len);
        for(int i=1;i<=c.len;i++){
            c.a[i]+=a[i]-b.a[i];
            if(c.a[i]<0){
                c.a[i]+=10;
                a[i+1]--;
            }
        }
        while(c.a[c.len+1])c.len++;
        while(c.len>1&&!c.a[c.len])c.len--;
        return c;
    }
    bool operator < (const bignum & b)const{
        if(b.len>len)return 1;
        for(int i=len;i;i--){
            if(b.a[i]>a[i])return 1;
            if(b.a[i]<a[i])return 0;
        }
        return 0;
    }
}A,B,C,tmpl,tmpr,tmpm,one;
int T,la,lb;
char sa[205],sb[205];
int main(){
    scanf("%d",&T);
    one=1;
    while(T--){
        scanf("%s%s",sa+1,sb+1);
        la=strlen(sa+1);lb=strlen(sb+1);
        memset(A.a,0,sizeof A.a);
        memset(B.a,0,sizeof B.a);
        A.len=la;B.len=lb;
        for(int i=1;i<=la;i++)A.a[la-i+1]=sa[i]-'0';
        for(int i=1;i<=lb;i++)B.a[lb-i+1]=sb[i]-'0';
        if(A.a[1]%2==0&&B.a[1]%2==0){puts("No");continue;}
        while(A.a[1]%2==0)A.div();
        while(B.a[1]%2==0)B.div();
        while(1){
            if(A<B){C=A;A=B;B=C;}
            if(B.len==1&&B.a[1]==0)break;
            A=A-B;
            while((A.len!=1||A.a[1]!=0)&&A.a[1]%2==0)A.div();
        }
        if(A.len==1&&A.a[1]==1)puts("Yes");
        else puts("No");
    }
    return 0;
}

T2:数数题，012的序列，问有多少个区间满足区间内最大的数不超过一半

正难则反，考虑有多少个不满足，不符合条件的区间为$j+1～i$,则一定有

$sumi-sumj>(i-j)/2$

$2*sumi-i>2*sumj-j$树状数组啊，于是我就60分gg了

再考虑性质，每往后移动一位，上述的val波动只有可能是1或-1；

于是处理前缀和O(1)转移即可

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 5000050
#define LL long long
using namespace std;
LL ans;
char s[N];
int n,a[N],sum[3][N];
int c[2*N],ss;
int main(){
    scanf("%d",&n);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++){
        a[i]=s[i]-'0';
        sum[0][i]=sum[0][i-1];
        sum[1][i]=sum[1][i-1];
        sum[2][i]=sum[2][i-1];
        sum[a[i]][i]++;
    }
    ans=(LL)n*(n+1)/2;
    for(int t=0;t<3;t++){
        memset(c,0,sizeof c);
        c[n+5]++;ss=1;
        for(int i=1,val,last;i<=n;i++){
            val=2*sum[t][i]-i+n+5;
            last=2*sum[t][i-1]-i+1+n+5;
            if(last==val-1){ans-=ss;}
            if(last==val+1){ss-=c[last]+c[val];ans-=ss;}
            c[val]++;
            ss+=c[val];
        }
    }
    printf("%lld
",ans);
    return 0;
}

T3:codeforces 739E加强版

O(n3)dp很好想，40分到手，然后我就gg了

如果贪心的话肯定是都喂A，也都喂B，但是球可能不够

考虑给每个A球赋一个额外的权值cost，每选一个A球就给最终答案减去cost，二分这个cost，就可以保证最终的A球在范围内，同理可以二分B

O(n3)->O(nlog2)

好像卡eps

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define eps 1e-8
#define N 100500
using namespace std;
int n,na,nb;
double p[N],q[N],pq[N];
double ff;
int num1,num2;
bool judge(double x,double y){
    num1=0,num2=0;ff=0;
    for(int i=1;i<=n;i++){
        int o=0;
        double tmp1=p[i]-x,tmp2=q[i]-y,tmp3=pq[i]-x-y,last=ff;
        if(last+tmp1>ff-eps){ff=last+tmp1;o=1;}
        if(last+tmp2>ff-eps){ff=last+tmp2;o=2;}
        if(last+tmp3>ff-eps){ff=last+tmp3;o=3;}
        if(o&1)num1++;if(o&2)num2++;
    }
}
double l,r,mid;
void check(double x){
    l=0;r=1;
    while(l+eps<r){
        mid=(l+r)/2.0;
        judge(x,mid);
        if(num2<=nb)r=mid;
        else l=mid;
    }
}
int main(){
    while(scanf("%d%d%d",&n,&na,&nb)==3){
        for(int i=1;i<=n;i++)scanf("%lf",&p[i]);
        for(int i=1;i<=n;i++)scanf("%lf",&q[i]);
        for(int i=1;i<=n;i++)pq[i]=p[i]+q[i]-p[i]*q[i];
        double L=0,R=1,MID;
        while(L+eps<R){
            MID=(L+R)/2.0;
            check(MID);
            judge(MID,r);
            if(num1<=na)R=MID;
            else L=MID;
        }
        judge(R,r);
        printf("%0.3f
",ff+R*na+r*nb);
    }
    return 0;
}