Codeforces Beta Round #107(Div2)

B.Phone Numbers

思路:就是简单的结构体排序,只是这里有一个技巧,就是结构体存储的时候,直接存各种类型的电话的数量是多少就行,在读入电话的时候,既然号码是一定的,那么就直接按照格式%c读取就好,在比较的时候也容易比较,直接比较ASCII码就行

代码:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
struct people{
    int num;
    char s[100];
    int t=0,p=0,c=0;
}a[1500];
bool cmpt(people a,people b){
    if(a.t==b.t){
        return a.num<b.num;
    }
    return a.t>b.t;
}
bool cmpp(people a,people b){
    if(a.p==b.p){
        return a.num<b.num;
    }
    return a.p>b.p;
}
bool cmpc(people a,people b){
    if(a.c==b.c){
        return a.num<b.num;
    }
    return a.c>b.c;
}
int main(){
    int t;
    scanf("%d",&t);
    for(int i=0;i<t;i++){
        a[i].num=i;
        int n;
        scanf("%d %s",&n,&a[i].s);
        int t=0;
        int p=0;
        int cc=0;

        for(int j=0;j<n;j++){
            char b,c,d,e,f,g;
            char s;
            getchar();
            scanf("%c%c",&b,&c);
            scanf("%c",&s);
            scanf("%c%c",&d,&e);
            scanf("%c",&s);
            scanf("%c%c",&f,&g);
            if(b==c&&c==d&&d==e&&e==f&&f==g){
                t++;
            }else if(b>c&&c>d&&d>e&&e>f&&f>g){
                p++;
            }else{
                cc++;
            }
        }
        a[i].p=p;
        a[i].t=t;
        a[i].c=cc;
    }
    /*for(int i=0;i<t;i++){
        printf("%d %d %d\n",a[i].t,a[i].p,a[i].c);
    }*/
    int tt=0;
    int p=0;
    int c=0;
    int sump=0,sumt=0,sumc=0;
    sort(a,a+t,cmpt);
    tt=a[0].t;
    for(int i=0;i<t;i++){
        if(a[i].t==tt){
            sumt++;
        }
    }
    printf("If you want to call a taxi, you should call: ");
    for(int i=0;i<sumt;i++){
        printf("%s",a[i].s);
        if(i<sumt-1){
            printf(", ");
        }
    }
    printf(".\n");
    sort(a,a+t,cmpp);
    p=a[0].p;
    for(int i=0;i<t;i++){
        if(a[i].p==p){
            sump++;
        }
    }
    printf("If you want to order a pizza, you should call: ");
    for(int i=0;i<sump;i++){
        printf("%s",a[i].s);
        if(i<sump-1){
            printf(", ");
        }
    }
    printf(".\n");
    sort(a,a+t,cmpc);
    c=a[0].c;
    for(int i=0;i<t;i++){
        if(a[i].c==c){
            sumc++;
        }
    }
    printf("If you want to go to a cafe with a wonderful girl, you should call: ");
    for(int i=0;i<sumc;i++){
        printf("%s",a[i].s);
        if(i<sumc-1){
            printf(", ");
        }
    }
    printf(".");
}

C.Win or Freeze

思路:其实最多两局,如果给定的这个数质因子大于2或者是质数,那么就是1赢,1直接选定两个质因子的乘积就能确保自己赢,剩下的情况就是2赢

代码:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<vector>
using namespace std;
vector <pair<__int64, __int64> > divide(__int64 x) {
    vector <pair<__int64, __int64> > v;
    for (int i = 2; i <= x / i; ++i)
        if (x % i == 0) {
            int s = 0;
            while (x % i == 0) x /= i, s++;
            v.emplace_back(i, s);
        }
    if (x > 1) v.emplace_back(x, 1);
    return v;
}

/*vector<__int64> get_divisors(__int64 x) {
    vector<__int64> res;
    for (__int64 i = 1; i <= x / i; ++i)
        if (x % i == 0) {
            res.push_back(i);
            if (i != x / i) res.push_back(x / i);
        }
    sort(res.begin(), res.end());
    return res;
}*/
inline bool isprime(long long n) {
    if (n < 2) return false;
    if (n == 2) return true;
    if (n % 2 == 0) return false;
    for (long long i = 3; i <= n / i; i += 2)
        if (n % i == 0) return false;
    return true;
}

int main(){
    __int64 num;
    int flag=0;
    scanf("%I64d",&num);
    if(num==1){
        printf("1\n");
        printf("0\n");
        return 0;
    }else if(isprime(num)==true){
        printf("1\n");
        printf("0\n");
        return 0;
    }
    vector<pair<__int64,__int64>> v;
    v=divide(num);
    __int64 len=v.size();
    __int64 sum=0;
    __int64 sump=1;
    int f=0;
    for(__int64 i=0;i<len;i++){
        __int64 se=v.back().second;
        __int64 fi=v.back().first;
        v.pop_back();
        sum+=se;
        if(f<2){
            if(se>=2&&f==0){
                f++;
                f++;
                sump*=fi;
                sump*=fi;
            }else if(se<2){
                sump*=fi;
                f++;
            }else{
                sump*=fi;
                f++;
            }
        }
    }
    if(sum<=2){
        printf("2\n");
    }else{
        printf("1\n");
        printf("%I64d\n",sump);
    }
}

D - Quantity of Strings

思路:主要分为k>n,k=n,k<n的情况,见代码,wa掉这么多主要是前两种考虑出了问题

代码:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<vector>
using namespace std;
const long long int modd=1e9+7;
int main(){
    int n,m,k;
    scanf("%d %d %d",&n,&m,&k);
    __int64 sum=0;
    if(k<n){
        if(k%2==0){
            sum+=m;
           /* if(k!=2){
                sum+=((m-1)*m)%modd;
            }*/
    }else{
        if(k==1){
            __int64 num=1;
            for(int i=0;i<n;i++){
                num=(num*m)%modd;
                num%=modd;
            }
            sum+=num;
        }else{
            __int64 num=1;

            sum+=((m-1)*m)%modd;

            sum+=m;
        }
    }
    }else if(k==n){
        if(k%2==0){
            __int64 num=1;
            for(int i=0;i<n/2;i++){
                num=(num*m)%modd;
                num%=modd;
            }
            sum+=num;
        }else{
            __int64 num=1;
            for(int i=0;i<(n/2)+1;i++){
                num=(num*m)%modd;
                num%=modd;
            }
            sum+=num;
        }
    }else{
        __int64 num=1;
            for(int i=0;i<n;i++){
                num=(num*m)%modd;
                num%=modd;
            }
            sum+=num;
    }
    printf("%I64d\n",sum);
}

 

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