Codeforces Round #174 (Div. 2)

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A题写麻烦了直接乘 p-1次就可以我还用快速幂把1--p-1都求了出来边求边取模

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stdlib.h>#include<cmath>using namespace std;int result(int a,int b,int m){    int d,t;    d=1;    t=a;    while (b>0)    {        if (b%2==1)            d=(d*t)%m;        b/=2;        t=(t*t)%m;    }    return d;}int main(){    int i,j,k,n,m,sum=0,flag = 1,s,g;    cin>>n;    for(i = 1; i < n ; i++)    {        flag  =1;        for(j = 1 ; j <= n-2 ; j++)        {            s = result(i,j,n);            if((s-1)%n==0)            {                flag = 0;                break;            }        }        if(!flag)        continue;        s = result(i,n-1,n);        if((s-1)%n!=0)        {            flag = 0;            continue;        }        sum++;    }    cout<<sum<<endl;    return 0;}

B题

#include<cstdio>#include<iostream>#include<cstring>#include<stdlib.h>using namespace std;char str[1000010];int main(){    int i,j=0,k,n,a=0,f=0;    char c;    cin>>n;    getchar();    for(i = 1; i <= n ; i++)    {        scanf("%c",&c);        if(c=='A')            a++;        else if(c=='F')            f++;        else            if(c=='I')                j++;    }    if(j==1)        cout<<"1"<<endl;    else        if(j==0)        {            cout<<a<<endl;        }        else            cout<<"0\n";    return 0;}

C题写挂了刚开始写了一半线段树看到好多人过觉得没那么难就随便写了个当时过了之后 WA了听队长说了简单方法

用一数组不断更新尾部应加的值若删掉这个数就把这个数值加到它的前一个注意删掉之后都更新为0

#include <iostream>#include<cstdio>#include<cstring>#include<stdlib.h>using namespace std;long long sum[1000010],x[1000010];int main(){    int i,j,k,n,m,a,b,c,g=1;    x[g] = 0;    long long s=0;    cin>>n;    while(n--)    {        cin>>a;        if(a==1)        {            cin>>b>>c;            sum[b] += c;            s+=(b*c);            printf("%.6lf\n",s*1.0/g);        }        else if(a==2)        {            cin>>b;            g++;            x[g] = b;            s+=b;            printf("%.6lf\n",s*1.0/g);        }        else        {            if(g>=2)            {                s-=(x[g]+sum[g]);                sum[g-1]+=sum[g];                sum[g] = 0;                x[g] = 0;                g--;                printf("%.6lf\n",s*1.0/g);            }        }    }    return 0;}