T1

#include <iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;

    int a = 1;//初始化，0刀是一块
    for(int i = 1; i <= n; i++)
    {
        //切第i刀时会多i块
        a += i;
    }

    cout << a;

    return 0;
}

T2

#include <iostream>

using namespace std;

int main()
{
    string s;
    cin >> s;
    int len = s.size();
    for(int i = 0; i < len - 4; i++)
    {
        if('b' == s[i] && 'l' == s[i + 1] && 'a' == s[i + 2] && 'c' == s[i + 3] && 'k' == s[i + 4])
        {
            s[i + 2] = 'o';
        }
    }

    cout << s;

    return 0;
}

T3

解法一

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int n;
    cin >> n;
    for(int i = -n; i <= n; i++)
    {
        cout << "**";

        int x = abs(i);
        if(0 == x)  //中间那行特判
        {
            cout << '*';
        }

        for(int j = 1; j <= x; j++)
        {
            cout << ' ';
        }
        for(int j = 1; j <= x; j++)
        {
            cout << '*';
        }
        cout << endl;
    }

    return 0;
}

解法二：Zirui代码

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
	int n;
	cin>>n;
	int tmp=n;
	for ( int i=0; i<n*2+1; i++ )
	{
		cout << "**";
		for ( int j=0; j<abs(tmp); j++ )
		{
			cout << " ";
		}
		if ( tmp==0 ) //中间行加空格
		{
			cout << "*";
		}
		for ( int j=0; j<abs(tmp); j++ )
		{
			cout << "*";
		}
		tmp--;
		cout << endl;
	}

	return 0;
 }

T4

解法一

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

const int maxN = 100000 + 5;
long long p[maxN];

int main()
{
    //freopen("T4.in", "r", stdin);

    int n, m;
    long long c;
    cin >> n >> m;
    long long total = 0;
    for(int i = 1; i <= n; i++)
    {
        cin >> p[i] >> c;
        if(c > 0)
        {
            //j必须在保质期内
            for(int j = max(1, i - m); j <= i; j++)
            {

                if(p[j] >= c)//第j天的牛奶满足需求，且有剩余
                {
                    total += c;
                    p[j] -= c;
                    c = 0;
                    break;
                }
                else    //第j天的牛奶少于需求，则下一天的牛奶也要卖
                {
                    total += p[j];
                    c -= p[j];
                    p[j] = 0;
                }
            }
        }
    }

    cout << total;

    return 0;
}

解法二：Theo代码

#include <iostream>
#include <queue>
#define LL long long
using namespace std;

struct milk
{
    int amount;
    int produced;
};

int n, m;
deque<milk> que;

int main()
{
    cin >> n >> m;
    LL tot = 0;
    for(int i = 1; i <= n; i++)
    {
        int p, c;
        cin >> p >> c;
        que.push_back({p, i});
        while (!que.empty())
        {
            if(i - que.front().produced > m)
            {
                que.pop_front();
            }
            if(que.front().amount > c)
            {
                tot += c;
                que.front().amount -= c;
                break;
            }
            if(que.front().amount == c)
            {
                tot += c;
                que.pop_front();
                break;
            }
            if(que.front().amount < c)
            {
                tot += que.front().amount;
                c -= que.front().amount;
                que.pop_front();
            }
        }
    }
    cout << tot << endl;
    return 0;
}

T5

解法一：70分代码

#include <iostream>
using namespace std;

const int maxN = 100;
int n, m, k, sum[maxN][maxN];
char  a[maxN][maxN];

void input()
{
    cin >> n >> m >> k;
    for(int r=1; r<=n; r++)
    {
        for(int c=1; c<=m; c++)
        {
            cin >> a[r][c];
        }
    }
}

//求二维数组前缀和
void getPreSum()
{
	for(int r=1; r<=n; r++)
    {
        for(int c=1; c<=m; c++)
        {
            sum[r][c] = sum[r-1][c] + sum[r][c-1] - sum[r-1][c-1] + (a[r][c] == '.');
        }
    }
}

int getSum(int r1, int c1, int r2, int c2)
{
    return sum[r2][c2] - sum[r1-1][c2] - sum[r2][c1-1] + sum[r1-1][c1-1];
}

int main()
{
    //freopen("T5.in", "r", stdin);

    input();
    getPreSum();

    const int INF = 2100000000;
    int minArea = INF;

    for(int r1 = 1; r1 <= n; r1++)
    {
        for(int c1 = 1; c1 <= m; c1++)
        {
            for(int r2 = r1; r2 <= n; r2++)
            {
                for(int c2 = c1; c2 <= m; c2++)
                {
                    if(getSum(r1, c1, r2, c2) >= k && (r2 - r1 + 1) * (c2 - c1 + 1) < minArea)
                    {
                        minArea = (r2 - r1 + 1) * (c2 - c1 + 1);
                    }
                }
            }
        }
    }

    if(minArea == INF)
    {
        cout << "No Solution";
    }
    else
    {
        cout << minArea;
    }

    return 0;
}

解法二： 尺取法

#include<bits/stdc++.h>
using namespace std;
char x;
int n,m,num,need,ans = 1e9,a[1101][1101];

void getSectionSum(int l,int r)
{
	int emptySeats = 0, up = 1, down = 1;
	for(; up <= n; up++)
    {
		for(; down <= n && emptySeats < need; down++)
		{
		    emptySeats += a[down][r] - a[down][l - 1];
		}

		if(emptySeats < need)
        {
            break;
        }

        //down最后又自加了一次，所以down-up不用再+1
		ans = min(ans,(r - l + 1) * (down - up));
		emptySeats -= a[up][r] - a[up][l - 1];
	}
}

int main()
{
    //freopen("T5.in", "r", stdin);

    cin >> n >> m >> need;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            cin>>x;
            num += (x == '.');
            a[i][j] = a[i][j - 1] + (x == '.');
        }
    }

	if(num < need)
    {
        cout<<"No Solution"<<endl;
        return 0;
	}

    for(int l = 1; l <= n; l++)
    {
        for(int r = l; r <= m; r++)
        {
            getSectionSum(l, r);
        }
    }

    cout<<ans<<endl;

    return 0;
}

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