Codeforces_101498

A.map统计数量，更新最大值。

#include<bits/stdc++.h>
using namespace std;

int n;
map<int,int> mp;

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        mp.clear();
        cin >> n;
        int ans,maxx = 0;
        for(int i = 1;i <= n;i++)
        {
            string s;
            int x;
            cin >> s >> x;
            mp[x]++;
            if(maxx == mp[x])    ans = min(ans,x);
            else if(maxx < mp[x])
            {
                ans = x;
                maxx = mp[x];
            }
        }
        cout << ans << endl;
    }
    return 0;
}

---

B.统计b中每个字符个数，a串线性扫一遍，到没有的字符则停止。

#include<bits/stdc++.h>
using namespace std;

char a[100005],b[100005];
int cnt[128];

int main()
{
    ios::sync_with_stdio(0);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",a,b);
        memset(cnt,0,sizeof(cnt));
        int len1 = strlen(a),len2 = strlen(b);
        for(int i = 0;i < len2;i++) cnt[b[i]]++;
        int ans = 0;
        for(int i = 0;i < len1;i++)
        {
            if(!cnt[a[i]]) break;
            ans++;
            cnt[a[i]]--;
        }
        printf("%d
",ans);
    }
    return 0;
}

---

C.选最小值。

#include<bits/stdc++.h>
using namespace std;

int a,b,c;

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        cin >> a >> b >> c;
        if(a < b && a < c)  cout << "First" << endl;
        else if(b < a && b < c) cout << "Second" << endl;
        else    cout << "Third" << endl;
    }
    return 0;
}

---

D.2*C(a-1,b)，预处理阶乘逆元。

#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;

long long a[100005],fac[100005],inv[100005];

long long c(int n,int m)
{
    return fac[n]*inv[m]%MOD*inv[n-m]%MOD;
}

int main()
{
    inv[0] = 1;
    inv[1] = 1;
    for(int i = 2;i <= 100000;i++)  inv[i] = inv[MOD%i]*(MOD-MOD/i)%MOD;
    for(int i = 2;i <= 100000;i++)  inv[i] = inv[i-1]*inv[i]%MOD;
    fac[0] = 1;
    for(int i = 1;i <= 100000;i++)  fac[i] = fac[i-1]*i%MOD;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long x,y;
        scanf("%lld%lld",&x,&y);
        printf("%lld
",2*c(x-1,y)%MOD);
    }
    return 0;
}

---

E.流水线，n+k-1。

#include<bits/stdc++.h>
using namespace std;

int main()
{
    ios::sync_with_stdio(0);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long x,y;
        scanf("%lld%lld",&x,&y);
        printf("%lld
",x+y-1);
    }
    return 0;
}

---

F.预处理每个数下一个的位置，set储存位置模拟。

#include<bits/stdc++.h>
using namespace std;

int n,k,a[100005],ne[100005];

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> k;
        for(int i = 1;i <= n;i++)   cin >> a[i];
        map<int,int> mp;
        set<int> s;
        for(int i = n;i >= 1;i--)
        {
            if(mp.count(a[i]))  ne[i] = mp[a[i]];
            else    ne[i] = n+i;
            mp[a[i]] = i;
        }
        int ans = 0;
        for(int i = 1;i <= n;i++)
        {
            if(s.count(i))
            {
                s.erase(i);
                s.insert(ne[i]);
                continue;
            }
            if(s.size() == k)  s.erase(*s.rbegin());
            s.insert(ne[i]);
            ans++;
        }
        cout << ans << endl;
    }
    return 0;
}

---

G.n^2枚举区间，只要sum%lcm则符合要求，lcm过大时，break掉。

#include<bits/stdc++.h>
using namespace std;

int n,a[2005];

long long lcm(long long a,long long b)
{
    long long t = __gcd(a,b);
    if(1e14/b > a/t)    return a/t*b;
    return 1e18;
}

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n;
        for(int i = 1;i <= n;i++)   cin >> a[i];
        int ans = 0;
        for(int i = 1;i <= n;i++)
        {
            long long t = 1,sum = 0;
            for(int j = i;j <= n;j++)
            {
                sum += a[j];
                t = lcm(t,a[j]);
                if(t > 1e14)    break;
                if(sum%t == 0)  ans++;
            }
        }
        cout << ans << endl;
    }
    return 0;
}

---

H.贪心，注意几种特例。

#include<bits/stdc++.h>
using namespace std;

int n,k;
char ans[1000005];

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> k;
        ans[n] = 0;
        if(n%2 == 0 && k%2 == 1 || 9*n < k || n > 1 && k < 2)
        {
            cout << -1 << endl;
            continue;
        }
        for(int i = 0,j = n-1;i < j;i++,j--)
        {
            if(k >= 18)
            {
                k -= 18;
                ans[i] = '9';
                ans[j] = '9';
            }
            else
            {
                int t = k/2;
                k -= 2*t;
                ans[i] = t+'0';
                ans[j] = t+'0';
            }
        }
        if(n%2) ans[n/2] = k+'0';
        cout << ans << endl;

    }
    return 0;
}

---

I.n,m均为偶数，先手必输。

#include<bits/stdc++.h>
using namespace std;

int n,m;

int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> m;
        if(n % 2 == 0 && m%2 == 0)  cout << "abdullah" << endl;
        else    cout << "hasan" << endl;
    }
    return 0;
}

---

J.从大到小枚举长度每一约数，当前约数不成立则把这个约数的所有因子标记不成立。

#include<bits/stdc++.h>
using namespace std;

char a[100005];
int vis[100005];

int main()
{
    ios::sync_with_stdio(0);
    int T;
    scanf("%d",&T);
    getchar();
    while(T--)
    {
        gets(a+1);
        int n = strlen(a+1);
        n++;
        a[n] = ' ';
        memset(vis,0,sizeof(vis));
        int flag = 0;
        for(int i = n/2;i >= 2;i--)
        {
            if(vis[i])  continue;
            if(n%i) continue;
            int ok = 1;
            for(int j = i;j <= n;j += i)
            {
                if(a[j] != ' ') ok = 0;
            }
            if(ok == 0)
            {
                for(int t = 2;t*t <= i;t++)
                {
                    if(i%t == 0)
                    {
                        vis[t] = 1;
                        vis[i/t] = 1;
                    }
                }
            }
            else
            {
                flag = 1;
                break;
            }
        }
        if(flag)    printf("YES
");
        else    printf("NO
");
    }
    return 0;
}

---