Codeforces Round #258 (Div. 2)

A

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<list>
using namespace std;
#define N 135
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


int main()
{

    int n,m;
    cin>>n>>m;
    if(min(n,m)%2)
    puts("Akshat");
    else puts("Malvika");
    return 0;
}

B

排序后比较

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<list>
using namespace std;
#define N 101000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
int a[N],b[N];

int main()
{

    int n,i;
    cin>>n;
    for(i = 1; i <= n; i++)
    {
        cin>>a[i];
        b[i] = a[i];
    }
    sort(a+1,a+n+1);
    int f = 0;
    int flag = 1,st = 1,en = 1;
    for(i = 1; i <= n ;i++)
    {
        if(a[i]!=b[i]&&!f)
        {
            st = i;
            f = 1;
        }
        else if(a[i]!=b[i])
        en = i;
    }
    int j = en;
    for(i = st ; i <= en ; i++)
    {
        if(a[j--]!=b[i])
        {
            flag = 0;
            break;
        }
    }
    if(flag) printf("yes
%d %d
",st,en);
    else printf("no
");
    return 0;
}

C

根据d1d2的符号 分四种情况讨论，注意k和n的取值范围

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<list>
using namespace std;
#define N 101000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
int a[N],b[N];

int main()
{

    int t,i,j;
    cin>>t;
    LL n,k,d1,d2;
    while(t--)
    {
        LL tk;
        scanf("%I64d%I64d%I64d%I64d",&n,&k,&d1,&d2);

        tk = n-k;
        if(n%3!=0)
        {
            puts("no");
            continue;
        }
        if(d1+d2+d2<=k&&3*(d1+d2)<=n)
        {
            LL k1 = d1+d2+d1;
            if((tk>=k1&&(tk-k1)%3==0))
            {
                puts("yes");
                continue;
            }
        }
        if(max(d1,d2)+abs(d1-d2)<=k&&3*d1<=n&&3*d2<=n)
        {
            LL k2 = d1+d2;
            if(tk>=k2&&(tk-k2)%3==0)
            {
                puts("yes");
                continue;
            }
        }
        if((d1+d2+d1)<=k&&3*(d2+d1)<=n)
        {
            LL k3 = d2+d2+d1;
            if((tk>=k3&&(tk-k3)%3==0))
            {
                puts("yes");
                continue;
            }
        }
        if((d1+d2)<=k&&3*max(d1,d2)<=n)
        {
            LL k4 = max(d1,d2)+abs(d2-d1);
            if(tk>=k4&&(tk-k4)%3==0)
            {
                puts("yes");
                continue;
            }
        }
        puts("no");
    }
    return 0;
}

D

因为只含ab 所以只要头尾相同肯定就是回文串，然后就统计一下第奇偶个位置上a,b的数目计数就可以了。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
char s[N];
LL num[2][2],ans[2];
int main()
{
    int n,i,j;
    cin>>s;
    int k = strlen(s);
    for(i = 0 ;i < k ;i++)
    {
        j = s[i]-'a';
        num[i&1][j]++;
        ans[0]+=num[(i+1)&1][j];
        ans[1]+=num[i&1][j];
    }
    cout<<ans[0]<<" "<<ans[1]<<endl;
    return 0;
}

E

组合数学的知识。

组合数学P113有同等类型题目讲解。

具有重复的组合，可以先求出无限制的T*集合  然后根据容斥原理去除超出限制的组合个数。

T* = C（n+m-1,n)  = C(n+m-1,m-1)

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
#define mod 1000000007
LL a[22],inv[22];
LL getInv(LL x)
{
    LL ret = 1;
    x %= mod;
    for (int a = mod - 2; a; a /= 2, x = x * x % mod)
        if (a % 2 == 1)
            ret = ret * x % mod;
    return ret;
}

int C(LL n, int m)
{
    if (n < m)
        return 0;
    int ret = 1;
    for (LL i = n; i > n - m; -- i)
        ret = ret * (i % mod) % mod * inv[n - i + 1] % mod;
    return ret;
}
int main()
{
    int n,i,j;
    LL s;
    for(i = 0 ; i <= 20 ; i++)
        inv[i] = getInv(i);
    cin>>n>>s;
    for(i = 0 ; i< n; i++)
        cin>>a[i];
    LL ans = C(s+n-1,n-1);//元素个数无限制的S集
    for(i = 1; i < (1<<n) ; i++)
    {
        LL ss = s;
        int o = 0;
        for(j =0 ; j< n ; j++)
            if(i&(1<<j))
            {
                ss-=(a[j]+1);//超过限制的
                o++;
            }
        if(ss<0) continue;
        if(o&1) ans-=C(ss+n-1,n-1);
        else ans+=C(ss+n-1,n-1);
        ans = (ans%mod+mod)%mod;//mod
    }
    cout<<ans<<endl;
    return 0;
}