Codeforces_835

A.比较两人总时间。

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

int s,v1,v2,t1,t2;

int main()
{
    ios::sync_with_stdio(false);
    cin >> s >> v1 >> v2 >> t1 >> t2;
    int x1 = s*v1+*t1,x2 = s*v2+*t2;
    if(x1 < x2) cout << "First" << endl;
    else if(x1 > x2)    cout << "Second" << endl;
    else    cout << "Friendship" << endl;
    return ;
}

---

B.记录每个数值个9的差的个数，贪心大的。

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

int k,cnt[];
string s;

int main()
{
    ios::sync_with_stdio(false);
    cin >> k >> s;
    int sum = ;
    for(int i = ;i < s.length();i++)
    {
        sum += s[i]-'';
        cnt[-s[i]+'']++;
    }
    int ans = ,now = ;
    while(sum < k)
    {
        while(cnt[now] == )    now--;
        cnt[now]--;
        sum += now;
        ans++;
    }
    cout << ans << endl;
    return ;
}

---

C.给周期内每个时间点都开数组，二维前缀和。

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

int n,q,c,ans[][][] = {};

int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> q >> c;
    for(int i = ;i <= n;i++)
    {
        int x,y,z;
        cin >> x >> y >> z;
        for(int j = ;j <= c;j++)   ans[x][y][j] += (j+z)%(c+);
    }
    for(int k = ;k <= c;k++)
    {
        for(int i = ;i <= ;i++)
        {
            for(int j = ;j <= ;j++) ans[i][j][k] += ans[i][j-][k]+ans[i-][j][k]-ans[i-][j-][k];
        }
    }
    while(q--)
    {
        int t,x1,x2,y1,y2;
        cin >> t >> x1 >> y1 >> x2 >> y2;
        t %= (c+);
        cout << ans[x2][y2][t]-ans[x1-][y2][t]-ans[x2][y1-][t]+ans[x1-][y1-][t] << endl;
    }
    return ;
}

---

D.区间dp。

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

string s;
int dp[][] = {},ans[] = {};

int main()
{
    ios::sync_with_stdio();
    cin >> s;
    s = ' '+s;
    for(int i = ;i < s.length();i++)   dp[i][i] = ;
    for(int i = ;i < s.length();i++)
    {
        if(s[i-] == s[i])  dp[i-][i] = ;
    }
    for(int len = ;len < s.length();len++)
    {
        for(int l = ;l+len- < s.length();l++)
        {
            int r = l+len-;
            if(s[l] != s[r] || !dp[l+][r-])    continue;
            dp[l][r] = dp[l][l+len/-]+;
        }
    }
    for(int i = ;i < s.length();i++)
    {
        for(int j = i;j < s.length();j++)   ans[dp[i][j]]++;
    }
    for(int i = s.length()-;i >= ;i--)    ans[i] += ans[i+];
    for(int i = ;i < s.length();i++)   cout << ans[i] << " ";
    cout << endl;
    return ;
}

---

E.因为有两个y，先把两个y划分进不同的组，把n个位置按位运算。每一位对应的值异或，统计只含一个y的位，10次。任取其中一位，对n个数划分成2块，之后对某一块少的二分找y的位置，9次。因为已经所有只含一个y的位，异或可得另一个y的位置。

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

int n,x,y;

int ask(vector<int> v)
{
    if(v.empty())   return ;
    cout << "? " << v.size();
    for(int i = ;i < v.size();i++) cout << " " << v[i];
    cout << endl;
    int x;
    cin >> x;
    return x;
}

int main()
{
    ios::sync_with_stdio();
    cin >> n >> x >> y;
    int diff = ,diffb;
    for(int i = ;i < ;i++)
    {
        vector<int> v;
        for(int j = ;j <= n;j++)
        {
            if(j&(<<i))    v.push_back(j);
        }
        int t = ask(v);
        if(t == y || t == (x^y))
        {
            diff |= (<<i);
            diffb = i;
        }
    }
    vector<int> a,b;
    for(int i = ;i <= n;i++)
    {
        if(i&(<<diffb))    a.push_back(i);
        else    b.push_back(i);
    }
    if(a.size() > b.size()) swap(a,b);
    int l = ,r = a.size()-;
    while(l < r)
    {
        int mid = (l+r)/;
        vector<int> v;
        for(int i = l;i <= mid;i++) v.push_back(a[i]);
        int t = ask(v);
        if(t == y ||t == (x^y)) r = mid;
        else    l = mid+;
    }
    int ans1 = a[l],ans2 = ans1^diff;
    if(ans1 > ans2) swap(ans1,ans2);
    cout << "! " << ans1 << " " << ans2 << endl;
    return ;
}