咕咕咕以後一定好好看題（

慘痛的教訓（

T1

因為根之間相差大於等於1，所以我們可以列舉長度為1的區間，然後就是浮點數的二分查詢，設個eps精度限制就行

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace
 
 std;
int cnt;
double a,b,c,d;
double eps = 0.001;
inline double f(double x){
    return x * x * x * a + x * x * b + x * c + d;    
}
int main(){
    freopen("equ.in","r",stdin);
    freopen("equ.out","w",stdout);
    scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
    //printf("%lf %lf %lf\n",f(0),f(2),f(100));
 

    for(int i=-100;i<=100 ;++i){
        double l = i,r = i + 1;
        double x1 = f(l),x2 = f(r);
        if(x1 == 0.0){
            cnt++;
            printf("%.2lf ",l);
            //continue;
        }
        if(x1 * x2 < 0){
            while(l + eps <= r){
                double mid = (l + r) / 2
 
;
                if(f(mid) * f(r) <= 0)l = mid;
                else r = mid;     
            }
            printf("%.2lf ",r);
            cnt++;
        }
        if(cnt == 3)break;
    }
}
//1 -5 -4 20

T2

 找規律我們可以發現，這個東西中心對稱，並且 i+2^k的區間都是由i的區間加上2^k過來的，模擬就行

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
typedef int ll;
ll n,m; 
int a[1005][1005];

int main(){
    freopen("match.in","r",stdin);
    freopen("match.out","w",stdout);
    scanf("%d",&m);
    n = (1 << m);
    int now = 1;
    a[1][1] = 1;
    int mm = m;
    while(m--){
        for(int i=1;i<=now;++i)
            for(int j=1;j<=now;++j)
                a[i][j+now] = a[i][j] + now;    
        for(int i=1;i<=now;++i)
            for(int j=1;j<=now;++j)
                a[i+now][j] = a[i][j+now],a[i+now][j+now] = a[i][j];
        now = now << 1;
    }
    for(int i=1;i<=n;++i){
        for(int j=1;j<=n;++j)printf("%d ",a[i][j]);
        putchar('\n');    
    }
}

 快速冪

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define int long long  
using namespace std;
typedef long long ll;
inline int qmid(int  a,int  b,int c){
    int res = 1,base = b;
    while(a){
        if(a&1)res = res * base % c;
        base = base * base % c;
        a >>= 1;    
    }
    return res;
}
inline int read(){
    register int x=0,f=1;
    register char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;
}
signed main(){
    freopen("mod.in","r",stdin);
    freopen("mod.out","w",stdout);
    int a,b,p;
    a=read(),b=read(),p=read();
    int x = qmid(b,a,p);
    printf("%lld^%lld mod %lld=%lld",a,b,p,x);
}

 沒看見step->痛失100分 兩位數沒有空格->痛失60分 多打一個空格->再次痛失60分

可以發現，除了最後四步，都是把o *最右邊的交換，最後4步的話就是這麼走的（可以這麼理解）然後就做完了

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
#define hahaha
using namespace std;
string s1[4] = {"ooo*o**--*", "o--*o**oo*", "o*o*o*--o*", "--o*o*o*o*"};
char s[300];
int n; 
int cnt;
void print(){
    printf("step%2d:",cnt++);
    for(int i=0;i<n*2+2;++i)putchar(s[i]);
    putchar('\n');    
}
inline void f(int x,int y){
    swap(s[x],s[y]);
    swap(s[x+1],s[y+1]);
    print();    
}
int main(){
    #ifdef hahaha
    freopen("chessman.in","r",stdin);
    freopen("chessman.out","w",stdout);
    #endif
    scanf("%d",&n);
    //int cnt = 0;
    for(int i=0;i<n;++i)s[i] = 'o';
    for(int i=n;i<n*2;++i)s[i] = '*';
    s[n*2] = s[n*2+1] = '-';
    print();
    int len = n;
    while(1){
        f(len-1,len*2);
        len--;
        if(len == 3)break;
        f(len,len*2);
    }
    string s2;
    for(int i=0;i<n-4;++i)s2+="o*";
    for(int i=0;i<4;++i){
        printf("step%2d:",cnt++);
        cout << s1[i] << s2 << endl;
    }
}