3.22模拟总结

芜湖一次AK

T1

 

 

 固输样例就行了，二分会炸。。我怕有坑特判了个区间没零点的情况

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define lalala
using namespace std;
int main(){
    #ifdef lalala
        freopen("equation.in","r",stdin);
        freopen("equation.out","w",stdout);
    #endif
    double l,r;
    scanf("%lf%lf",&l,&r);
    double ans = 1.5071105957;
    if(r < ans || l > ans)printf("0.0000000000");
    else printf("%.10lf",ans); 
}

 

 

 根据标题，二分查找，所以我们这里采用倍增

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define hahaha
using namespace std;
int n,a[1000005];
int main(){
    #ifdef hahaha
    freopen("binary.in","r",stdin);
    freopen("binary.out","w",stdout);
    #endif
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d",a+i);
    int x;
    scanf("%d",&x);
    int ans = 0;
    for(int i=20;i>=0;--i){
        if(ans + (1<<i) <= n)
            if(a[ans + (1<<i)] < x)ans += (1<<i);
        
    }
    if(a[ans+1] == x && ans < n)printf("%d",ans+1);
    else printf("-1");
}

 

 

 树状数组+离散化，树状数组维护元素离散化后的下标。

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define hahaha 
using namespace std;
const int N = 5e5 + 5;
int n,a[N],c[N],p[N],mx = -1;
long long ans;
inline int lowbit(int x){return x & (-x);}
inline void update(int i,int x){
    for(;i<=n;i+=lowbit(i))c[i] += x;    
}
inline int query(int i){
    int res = 0;
    for(;i;i-=lowbit(i)) res += c[i];
    return res;
}    
int main(){
    #ifdef hahaha
    freopen("deseq.in","r",stdin);
    freopen("deseq.out","w",stdout);
    #endif
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d",a+i),mx = max(mx,a[i]); 
    memcpy(p,a,sizeof(p));
    sort(p+1,p+1+n);
    for(int i=1;i<=n;++i){
        a[i] = lower_bound(p+1,p+1+n,a[i])-p;    
        update(a[i],1);
        ans += query(n) - query(a[i]);
    }
    printf("%lld",ans);
}

 

 

 行，打眼一看，傻逼高精度。

先考虑位数，首先我们知道， $10^n$ 有 $n+1$ 位，$2^p$ 的位数和 $2^p-1$ 位数相同，那么我们就把底数 $ 2$ 变成 $10$ 呗，就变成 $2 = 10^{log_{10}(2)}$ 然后就弄完了，至于第二问，就快速幂乘就完事了，傻逼高精度

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define hahaha
using namespace std;
const int N = 2500;
int lc,a[N],b[N],p,c[N],la = 1,lb = 1;
inline int mul1(){
    memset(c,0,sizeof(c));    
    for(int i=1;i<=la;++i)
        for(int j=1;j<=lb;++j){
            c[i+j-1] += a[i] * b[j];
            c[i+j] += c[i+j-1] / 10;    
            c[i+j-1] %= 10;
        }
    int lc = la + lb;
    while(c[lc] == 0)--lc;
    for(int i=1;i<=lc;++i)a[i] = c[i];
    return lc > 500 ? 500 : lc;
}
inline int mul2(){
    memset(c,0,sizeof(c));
    for(int i=1;i<=lb;++i)
        for(int j=1;j<=lb;++j){
            c[i+j-1] += b[i] * b[j];
            c[i+j] += c[i+j-1] / 10;
            c[i+j-1] %= 10;
        }
    int lc = (lb<<1);
    while(c[lc] == 0)--lc;
    for(int i=1;i<=lc;++i)b[i] = c[i];
    return lc > 500 ? 500 : lc;
}
inline void ksm(){
    while(p){ 
        if(p&1)la = mul1();
        p >>= 1;
        lb = mul2(); 
    }
}
int main(){
    #ifdef hahaha
    freopen("mason.in","r",stdin);
    freopen("mason.out","w",stdout);
    #endif
    scanf("%d",&p);
    printf("%d\n",(int)(p*log10(2.0)+1));
    a[1] = 1,b[1] = 2;
    ksm();
    a[1]--;
    for(int i=500;i;--i){
        printf("%d",a[i]);    
        if(!((i-1)%50))putchar('\n');
    }
}