NOIP模拟测试38「金·斯诺·赤」

金

辗转相减见祖宗

高精

#include<bits/stdc++.h>
using namespace std;
#define A 2000
#define P 1
#define N 10
#define ll long long
ll n,T;
char sjdfj[A];
struct bignum
{ll n[A],l;bignum(){l=1,memset(n,0,sizeof(n));}void clear(){while(l>1&&!n[l-1]) l--;}void print(){printf("%lld",n[l-1]);for(ll i=l-2;i>=0;i--){printf("%0*lld",P,n[i]);}printf("\n");}void read(){l=0;scanf("%s",sjdfj+1);l=strlen(sjdfj+1);reverse(sjdfj+1,sjdfj+l+1);for(ll i=0;i<l;i++){n[i]=sjdfj[i+1]-'0';}}ll ok(){//若为0 return1 //若%2==0 return2//若%2!=0 return3if(n[0]==0&&l==1) return 1;
//        if(n[0]==1&&l==1) return 1;if(n[0]%2==0) return 2;if(n[0]%2!=0) return 3;}bool operator <(bignum x) const{bignum t=*this,tep;if(t.l!=x.l)    return t.l<x.l;for(ll i=t.l-1;i>=0;i--){if(t.n[i]!=x.n[i]) return t.n[i]<x.n[i];}return 0;}bool operator >(bignum x) const{bignum t=*this;if(t.l!=x.l) return t.l>x.l;for(ll i=t.l-1;i>=0;i--){if(t.n[i]!=x.n[i]) return t.n[i]>x.n[i];}return 0;}bignum operator -(bignum x) const{bignum t=*this;if(t<x) swap(t,x);ll jie=0;//    t.print();x.print();for(ll i=0;i<t.l;i++){t.n[i]-=x.n[i];while(t.n[i]<0){t.n[i]+=N;jie++;}t.n[i+1]-=jie;jie=0;}while(!t.n[t.l-1]&&t.l>1) t.l--;return t;}bignum operator *(bignum x) const{bignum t=*this,tep;tep.l=t.l+x.l+1;for(ll i=0;i<t.l;i++)for(ll j=0;j<x.l;i++){tep.n[i+j]+=t.n[i]*x.n[j];}for(ll i=0;i<tep.l;i++){if(tep.n[i]>=N) {tep.n[i+1]+=tep.n[i]/N;tep.n[i]%=N;}}tep.clear();return tep;}bignum operator +(bignum x)const{bignum t=*this;if(t.l<x.l) t.l=x.l;t.l++;for(ll i=0;i<t.l;i++){t.n[i]+=x.n[i];if(t.n[i]>=N){t.n[i+1]+=t.n[i]/N;t.n[i]%=N;}}t.clear();return t;}bignum operator =(ll x){l=0;while(x){n[l++]=x%N;x/=N;}return *this;}bignum operator *(const ll &b){bignum t=*this,r;r.l=0;ll g=0;for(ll i=0;i<t.l||g;i++){ll x;if(i<t.l)x=t.n[i]*b+g;else x=g;r.n[r.l++]=x%N;g=x/N;}return r;}bignum operator /(const ll &x){bignum t=*this,r;ll tmp=0;r.l=t.l;for(ll i=t.l-1;i>=0;i--){tmp+=t.n[i];if(tmp>=x){r.n[i]=tmp/x;tmp%=x;}tmp*=N;}r.clear();return r;}
}a,b,c;
ll gcd(){//若为0 return1 //若%2==0 return2//若%2!=0 return3while((a.ok()!=1&&b.ok()!=1)){
//        printf("a=%lld ",a.ok());
//        a.print();
//        printf("b=%lld ",b.ok());
//        b.print();ll ok1=a.ok(),ok2=b.ok();if(ok1==2&&ok2==2){return 0;a=a/2,b=b/2;}else if(ok1==3&&ok2==3){if(a<b) swap(a,b);a=a-b;}else if(ok1==2&&ok2==3){a=a/2;}else if(ok1==3&&ok2==2){b=b/2;}}
}
int main()
{/*1 1023 3072*/
//    freopen("bf.txt","w",stdout);scanf("%lld",&T);for(ll i=1;i<=T;i++){a.read(),b.read();//    a=a-b;
        gcd();
//        a.print(),b.print();if(a.n[0]==0&&b.n[0]==1&&b.l==1&&a.l==1){printf("Yes\n");}else if(a.n[0]==1&&b.n[0]==0&&b.l==1&&a.l==1){printf("Yes\n");}else printf("No\n");}
}

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斯诺

考试代码改了改,数组开小见祖宗

考试时候$re$了

大概就是这样

考试时也维护的前缀和

$60\%$算法

只含$0,1$

我们可以将$0$看作减$1$,$1$看作加一

那么合法方案数就是$sum[r]-sum[l-1]==0$的个数

我们开一个桶存$sum[l]$,那么当前符合就是桶里$sum[i]$个数

查完个数再把$sum[r]$压进桶就行了

注意初始化,当你$sum==0$时也是合法方案,方案数为桶里$sum==0$个数$+1$,你可以先在桶里$0$压一个再进行操作

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 5000000
ll tong[mod+mod+mod+mod],sum[mod+mod][3],sumtp[21111111],sum2[111111111];
ll n,ans=0,all;
char a[mod+mod];
void solve(ll ql,ll qr){if(ql==qr) return ;ll mid=(ql+qr)>>1;solve(ql,mid);solve(mid+1,qr);ans=rand();
}
ll check(ll l,ll r){ll len=(r-l+1)/2;
//    printf("l=%lld r=%lld len=%lld\n",l,r,len);
//    printf("sum0=%lld 1=%lld 2=%lld\n",sum[r][0]-sum[l-1][0],sum[r][1]-sum[l-1][1],sum[r][2]-sum[l-1][2]);if(sum[r][0]-sum[l-1][0]>len) return 0;if(sum[r][1]-sum[l-1][1]>len) return 0;if(sum[r][2]-sum[l-1][2]>len) return 0;return 1;
}
int main(){scanf("%lld",&n);scanf("%s",a+1);all=1;for(ll i=1;i<=n;i++){sum[i][0]=sum[i-1][0];sum[i][1]=sum[i-1][1];sum[i][2]=sum[i-1][2];if(a[i]=='0') sum[i][0]++;if(a[i]=='1') sum[i][1]++;if(a[i]=='2') sum[i][2]++,all=0;}if(all&&n>1000){tong[mod]=1;for(ll i=1;i<=n;i++){sumtp[i]=sumtp[i-1];if(a[i]=='0') {
//                if(a[i-1]=='1') sumtp[i]=0;
//                if(sumtp[i]<0) ans++;sumtp[i]++;}else {
//                if(a[i-1]=='0') sumtp[i]=0;
//                if(sumtp[i]>0) ans++;sumtp[i]--;}}for(ll i=1;i<=n;i++){ans+=tong[mod+sumtp[i]];tong[mod+sumtp[i]]++;
//            printf("sumtp=%lld\n",sumtp[i]);
            }printf("%lld\n",ans);return 0;}if(n<=1000)for(ll i=1;i<=n;i++){for(ll j=i+1;j<=n;j++){if(check(i,j)){ans++;}}}else solve(1,n);printf("%lld\n",ans);
}