Codeforces_837

A.扫一遍。

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

int n;
string s;

int main()
{
    cin >> n;
    getchar();
    getline(cin,s);
    int ans = 0,maxx = 0;
    for(int i = 0;i < n;i++)
    {
        if(s[i] == ' ') maxx = 0;
        else
        {
            if(s[i] >= 'A' && s[i] <= 'Z') maxx++;
            ans = max(ans,maxx);
        }
    }
    cout << ans << endl;
    return 0;
}

---

B.暴力判断两种情况。

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

int n,m;
string s[105],ss[105];

int main()
{
    cin >> n >> m;
    for(int i = 1;i <= n;i++)
    {
        cin >> s[i];
        s[i] = ' '+s[i];
    }
    int flag = 0;
    if(n%3 == 0)
    {
        int t1 = n/3+1,t2 = n/3*2+1;
        char a = s[1][1],b = s[t1][1],c = s[t2][1];
        if(a != b && b != c && a != c)
        {
            int ok = 1;
            for(int i = 1;i < t1;i++)
            {
                for(int j = 1;j <= m;j++)
                {
                    if(s[i][j] != a)    ok = 0;
                }
            }
            for(int i = t1;i < t2;i++)
            {
                for(int j = 1;j <= m;j++)
                {
                    if(s[i][j] != b)    ok = 0;
                }
            }
            for(int i = t2;i <= n;i++)
            {
                for(int j = 1;j <= m;j++)
                {
                    if(s[i][j] != c)    ok = 0;
                }
            }
            if(ok)  flag = 1;
        }
    }
    if(m%3 == 0)
    {
        int t1 = m/3+1,t2 = m/3*2+1;
        char a = s[1][1],b = s[1][t1],c = s[1][t2];
        if(a != b && b != c && a != c)
        {
            int ok = 1;
            for(int i = 1;i < t1;i++)
            {
                for(int j = 1;j <= n;j++)
                {
                    if(s[j][i] != a)    ok = 0;
                }
            }
            for(int i = t1;i < t2;i++)
            {
                for(int j = 1;j <= n;j++)
                {
                    if(s[j][i] != b)    ok = 0;
                }
            }
            for(int i = t2;i <= m;i++)
            {
                for(int j = 1;j <= n;j++)
                {
                    if(s[j][i] != c)    ok = 0;
                }
            }
            if(ok)  flag = 1;
        }
    }
    if(flag)    cout << "YES" << endl;
    else    cout << "NO" << endl;
    return 0;
}

---

C.暴力取两辆，判断是否可行。

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

int n,a,b,x[105],y[105];

bool ok(int i,int j)
{
    if(x[i]+y[j] <= a && max(y[i],x[j]) <= b)    return 1;
    if(x[i]+y[j] <= b && max(y[i],x[j]) <= a)    return 1;
    if(x[i]+x[j] <= a && max(y[i],y[j]) <= b)    return 1;
    if(x[i]+x[j] <= b && max(y[i],y[j]) <= a)    return 1;
    if(y[i]+y[j] <= b && max(x[i],x[j]) <= a)    return 1;
    if(y[i]+y[j] <= a && max(x[i],x[j]) <= b)    return 1;
    if(y[i]+x[j] <= b && max(x[i],y[j]) <= a)    return 1;
    if(y[i]+x[j] <= a && max(x[i],y[j]) <= b)    return 1;
    return 0;
}

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> a >> b;
    for(int i = 1;i <= n;i++)   cin >> x[i] >> y[i];
    int ans = 0;
    for(int i = 1;i <= n;i++)
    {
        for(int j = i+1;j <= n;j++)
        {
            if(ok(i,j)) ans = max(ans,x[i]*y[i]+x[j]*y[j]);
        }
    }
    cout << ans << endl;
    return 0;
}

---

D.dp。

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

int n,k,cnt1[205] = {0},cnt2[205] = {0},dp[205][6005];
long long a[205];

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> k;
    for(int i = 1;i <= n;i++)   cin >> a[i];
    for(int i = 1;i <= n;i++)
    {
        while(a[i]%2 == 0)
        {
            a[i] /= 2;
            cnt2[i]++;
        }
        while(a[i]%5 == 0)
        {
            a[i] /= 5;
            cnt1[i]++;
        }
    }
    memset(dp,-1,sizeof(dp));
    dp[0][0] = 0;
    for(int i = 1;i <= n;i++)
    {
        for(int j = k;j >= 1;j--)
        {
            for(int t = 6000;t >= cnt1[i];t--)
            {
                if(dp[j-1][t-cnt1[i]] != -1)    dp[j][t] = max(dp[j][t],dp[j-1][t-cnt1[i]]+cnt2[i]);
            }
        }
    }
    int ans = 0;
    for(int i = 0;i <= 6000;i++)    ans = max(ans,min(i,dp[k][i]));
    cout << ans << endl;
    return 0;
}

---

E.有很多次操作，只会把b减一，我们对于每一次gcd != 1之前，对a的每个因子取模统计 gcd == 1的次数，统一减。另外，已经被b约去的因子，可以在之后的步骤不考虑。

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

long long a,b;

int main()
{
    ios::sync_with_stdio(0);
    cin >> a >> b;
    long long t = __gcd(a,b);
    a /= t;
    b /= t;
    vector<long long> v;
    for(long long i = 2;i*i <= a;i++)
    {
        while(a%i == 0)
        {
            a /= i;
            v.push_back(i);
        }
    }
    if(a > 1)   v.push_back(a);
    long long ans = 0;
    while(b)
    {
        long long t = b;
        for(auto it = v.begin();it != v.end();it++) t = min(t,b%(*it));
        ans += t;
        b -= t;
        vector<long long> vv;
        for(auto it = v.begin();it != v.end();it++)
        {
            long long t = *it;
            if(b%t == 0)    b /= t;
            else    vv.push_back(t);
        }
        v = vv;
    }
    cout << ans << endl;
    return 0;
}

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