Codeforces Round #411 (Div. 2)

A题

分析:傻逼题，两个数的差不大于1输出任意一个，否则输出2

#include "iostream"
#include "cstdio"
#include "cstring"
#include "cmath"
using namespace std;
int main()
{
    long long a,b;
    while(cin>>a>>b)
    {
        if(abs(a-b)<=1)
            cout<<a<<endl;
        else
            cout<<"2"<<endl;
    }
    return 0;
}

B题

分析:还是傻逼题，只要输出aabb不断循环下去就行

#include "iostream"
#include "cstdio"
#include "cstring"
#include "cmath"
using namespace std;
int n;
int main()
{
    while(cin>>n)
    {
        int ans=n/4;
        string a="aabb";
        for(int i=1;i<=ans;i++)
            cout<<a;
        int cnt=n%4;
        for(int i=0;i<cnt;i++)
            cout<<a[i];
        cout<<endl;
    }
    return 0;
}

C题

分析:经推到为(n-1)/2

#include "iostream"
#include "cstdio"
#include "cstring"
using namespace std;
long long n;
int main()
{
    while(cin>>n)
        cout<<(n-1)/2<<endl;
}

D题

分析:最终都会转换为bbbaa...这样的形式，倒着推一下就有了

#include "iostream"
#include "cstdio"
#include "cstring"
#include "string"
#include "cmath"
using namespace std;
const int mod=1000000007;
const int maxn=100010;
string s;
long long vis[100010];
int main()
{
    cin>>s;
        int len=s.length();
        long long m,l,k,ans,numa,numb;
        numa=0;
        numb=0;
        ans=0;
        for(int i=len-1;i>=0;i--)
        {
            if(s[i]=='a')
                numa++;
            if(s[i]=='b')
                numb++;
            if(s[i]=='a' && numb>0)
            {
                ans=(ans+numb)%mod;
                numb=(numb*2)%mod;
            }
        }
        cout<<ans<<endl;
    return 0;
}