2020杭电多校第六场部分题解（1001/1002/1009）

1001

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6827

题意：给定n个数，求第i个数到第j个数的平均值的期望（ij为1到n的随机数且i<=j）

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
const int N=2e6+5;
typedef unsigned long long ull;
typedef long long ll;
ll quickpow(ll a, ll b) {
    if (b < 0) return 0;
    ll ret = 1;
    a %= mod;
    while(b) {
        if (b & 1) ret = (ret * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ret;
}
ll Inv(ll a) {
    return quickpow(a, mod - 2);
}
int n;
ll inv[N],J[N];
void init()
{
    inv[1] = 1;J[1]=1;
    for(ll i = 2; i < N; ++ i){
        inv[i] = (mod - mod / i) * inv[mod % i] % mod;
        J[i]=(J[i-1]+i)%mod;
    }
}
int a[N];
ll sum[N],pre[N];
int main()
{
    init();
    int t,i,j,n;scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        ll gs=n/2;if(n%2==1)gs++;
        memset(sum,0,sizeof sum);
        memset( pre,0,sizeof pre);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            sum[1]=(a[i]+sum[1])%mod;
        }
        pre[1]=sum[1];
        for(i=2;i<=gs;i++){
            pre[i]=(pre[i-1]-a[i-1]-a[n-i+2]+mod)%mod;
            sum[i]=(sum[i-1]+pre[i])%mod;
        }
        ll ans=0;
        for(i=1;i<=gs;i++){
            ans=(ans+sum[i]*inv[i]%mod)%mod;
        }
        int k=n%2==0?gs:gs-1;
        for(i=gs+1;i<=n;i++){
            ans=(ans+sum[k--]*inv[i]%mod)%mod;
        }
        printf("%lld\n",ans*Inv(J[n])%mod);
    }
    return 0;
}

1002

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6827

题意：给定一个等式，问在几进制下等式成立

思路：模拟

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    char s[25];
    while(~scanf("%s",s)){
        int maxn=1,l=strlen(s);
        char ch;

        for(int i=0;i<l;i++){
            if(s[i]>='0'&&s[i]<='9')maxn=max(maxn,s[i]-'0');
            else if(s[i]>='A'&&s[i]<='F')maxn=max(maxn,s[i]-'A'+10);
            if(s[i]=='+'||s[i]=='-'||s[i]=='*'||s[i]=='/') ch=s[i];
        }
        int flag=0;
        for(int i=maxn+1;i<=16;i++){
            ll a=0,b=0,c=0,ab=-1;
            int j=0;
            for(;j<l;j++){
                if(s[j]>='0'&&s[j]<='9') a=1LL*a*i+s[j]-'0';
                else if(s[j]>='A'&&s[j]<='F') a=1LL*a*i+s[j]-'A'+10;
                else break;
            }
            for(j=j+1;j<l;j++){
                if(s[j]>='0'&&s[j]<='9') b=1LL*b*i+s[j]-'0';
                else if(s[j]>='A'&&s[j]<='F') b=1LL*b*i+s[j]-'A'+10;
                else break;
            }
            for(j=j+1;j<l;j++){
                if(s[j]>='0'&&s[j]<='9') c=1LL*c*i+s[j]-'0';
                else if(s[j]>='A'&&s[j]<='F') c=1LL*c*i+s[j]-'A'+10;
            }
            if(ch=='+')ab=a+b;
            else if(ch=='-') ab=a-b;
            else if(ch=='*') ab=a*b;
            else if(ch=='/'){
                if(a%b==0) ab=a/b;
            }
            //printf("%lld %lld %lld %lld\n",a,b,ab,c);
            if(ab==c){
                printf("%d\n",i);
                flag=1;
                break;
            }
        }
        if(!flag) printf("-1\n");
    }
}

1009

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6827

题意：对于任意b进制数，若y的各位之和可以被x整除，则y也能被x整除，反之不能

官方题解：

#include<bits/stdc++.h>
using namespace std;
const int N=2e6+5;

int main()
{
    int t;
    long long b,x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&b,&x);
        if(b>x)
        {
            if(__gcd(b-1,x)==x)printf("T\n");
            else printf("F\n");
        }
        else printf("F\n");
    }
    return 0;
}