Codeforces_816

A.不断增加时间，直到符合要求。

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

int a,b;
char c;

int f(int x)
{
    return x%10*10+x/10;
}

int main()
{
    ios::sync_with_stdio(0);
    cin >> a >> c >> b;
    int ans = 0;
    while(1)
    {
        if(f(a) == b)   break;
        ans++;
        b++;
        if(b == 60)
        {
            b = 0;
            a++;
        }
        if(a == 24) a = 0;
    }
    cout << ans << endl;
    return 0;
}

---

B.两次前缀和。

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

int n,k,q,a[200005] = {0},ans[200005] = {0};

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> k >> q;
    for(int i = 1;i <= n;i++)
    {
        int x,y;
        cin >> x >> y;
        a[x]++;
        a[y+1]--;
    }
    for(int i = 1;i <= 200000;i++)   a[i] += a[i-1];
    for(int i = 1;i <= 200000;i++)
    {
        ans[i] = ans[i-1];
        if(a[i] >= k)   ans[i]++;
    }
    while(q--)
    {
        int x,y;
        cin >> x >> y;
        cout << ans[y]-ans[x-1] << endl;
    }
    return 0;
}

---

C.对于每一行，我们找当前最小的值，然后处理该行每一个值，每一个值减去最小值，若结果不为0，则该列处理这个值至0，最后判断是否每个点都为0。

注意n>m的时候，我们转置一下矩阵才能得到最小的答案。

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

int n,m,a[105][105],cnt1[105] = {0},cnt2[105] = {0};

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> m;
    int flag = 1;
    if(n > m)   flag = 0;
    if(flag)
    {
        for(int i = 1;i <= n;i++)
        {
            for(int j = 1;j <= m;j++)   cin >> a[i][j];
        }
    }
    else
    {
        swap(n,m);
        for(int j = 1;j <= m;j++)
        {
            for(int i = 1;i <= n;i++)   cin >> a[i][j];
        }
    }
    for(int i = 1;i <= n;i++)
    {
        int minn = 1000;
        for(int j = 1;j <= m;j++)   minn = min(minn,a[i][j]);
        for(int j = 1;j <= m;j++)
        {
            if(a[i][j] != minn)
            {
                int t = a[i][j]-minn;
                cnt2[j] += t;
                for(int k = 1;k <= n;k++)   a[k][j] -= t;
            }
            a[i][j] -= minn;
        }
        cnt1[i] += minn;
    }
    int ok = 1;
    for(int i = 1;i <= n;i++)
    {
        for(int j = 1;j <= m;j++)
        {
            if(a[i][j] != 0)    ok = 0;
        }
    }
    if(!ok)
    {
        cout << -1 << endl;
        return 0;
    }
    int ans = 0;
    for(int i = 1;i <= n;i++)   ans += cnt1[i];
    for(int i = 1;i <= m;i++)   ans += cnt2[i];
    cout << ans << endl;
    if(flag)
    {
        for(int i = 1;i <= n;i++)
        {
            while(cnt1[i])
            {
                cnt1[i]--;
                cout << "row " << i << endl;
            }
        }
        for(int i = 1;i <= m;i++)
        {
            while(cnt2[i])
            {
                cnt2[i]--;
                cout << "col " << i << endl;
            }
        }
    }
    else
    {
        for(int i = 1;i <= n;i++)
        {
            while(cnt1[i])
            {
                cnt1[i]--;
                cout << "col " << i << endl;
            }
        }
        for(int i = 1;i <= m;i++)
        {
            while(cnt2[i])
            {
                cnt2[i]--;
                cout << "row " << i << endl;
            }
        }
    }
    return 0;
}

---

D.发现n是偶数的时候，会有规律，每隔两行会是前面的两个数相加，直接计算二项式的系数，最后一步的加减取决于n%4。

如果n是奇数，先模拟一行变成偶数。

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

int n,a[200005];
long long fac[100005] = {1};

long long qpow(long long a,long long b)
{
    long long ans = 1;
    while(b)
    {
        if(b%2) ans = ans*a%MOD;
        a = a*a%MOD;
        b /= 2;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(0);
    for(int i = 1;i <= 100000;i++)  fac[i] = fac[i-1]*i%MOD;
    cin >> n;
    for(int i = 0;i < n;i++)    cin >> a[i];
    if(n == 1)
    {
        cout << a[0] << endl;
        return 0;
    }
    if(n%2)
    {
        n--;
        int now = 1;
        for(int i = 0; i < n;i++)
        {
            if(now) a[i] = (a[i]+a[i+1])%MOD;
            else    a[i] = (a[i]-a[i+1]+MOD)%MOD;
            now ^= 1;
        }
    }
    n = n/2-1;
    long long ans1 = 0,ans2 = 0;
    for(int i = 0;i <= n;i++)
    {
        long long t = fac[n]*qpow(fac[i]*fac[n-i]%MOD,MOD-2)%MOD;
        ans1 = (ans1+t*a[i*2]%MOD)%MOD;
        ans2 = (ans2+t*a[i*2+1]%MOD)%MOD;
    }
    if(n%2) cout << (ans1-ans2+MOD)%MOD << endl;
    else    cout << (ans1+ans2)%MOD << endl;
    return 0;
}

---

E.树形DP，dp[i][j][k]代表(当前在i点，买了j个商品，是否购买当前商品)的最少花费。

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

int n,m,a[5005],b[5005],sz[5005] = {0},ok[5005] = {0};
long long dp[5005][5005][2];
vector<int> v[5005];

void dfs(int now)
{
    if(ok[now]) return;
    dp[now][0][0] = 0;
    dp[now][1][0] = a[now];
    dp[now][1][1] = b[now];
    sz[now] = 1;
    for(int i = 0;i < v[now].size();i++)
    {
        int t = v[now][i];
        dfs(t);
        for(int i = sz[now];i >= 0;i--)
        {
            for(int j = sz[t];j >= 0;j--)
            {
                dp[now][i+j][0] = min(dp[now][i+j][0],dp[now][i][0]+dp[t][j][0]);
                dp[now][i+j][1] = min(dp[now][i+j][1],dp[now][i][1]+min(dp[t][j][0],dp[t][j][1]));
            }
        }
        sz[now] += sz[t];
    }
    ok[now] = 1;
}

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> m;
    for(int i = 1;i <= n;i++)
    {
        cin >> a[i] >> b[i];
        b[i] = a[i]-b[i];
        if(i > 1)
        {
            int x;
            cin >> x;
            v[x].push_back(i);
        }
    }
    memset(dp,0x3f,sizeof(dp));
    dfs(1);
    for(int i = n;i >= 0;i--)
    {
        if(dp[1][i][0] <= m || dp[1][i][1] <= m)
        {
            cout << i << endl;
            return 0;
        }
    }
    return 0;
}