Codeforces Round 894 (Div. 3)

A.Gift Carpet

签到题

def solve():
    n, m = map(int, input().split())
    a = [list(input()) for _ in range(n)]
    flag = [False] * 4

    for j in range(m):
        for i in range(n):
            if not flag[0] and a[i][j] == 'v':
                flag[0] = True
                break
            elif flag[0] and not flag[1] and a[i][j] == 'i':
                flag[1] = True
                break
            elif flag[0] and flag[1] and not flag[2] and a[i][j] == 'k':
                flag[2] = True
                break
            elif flag[0] and flag[1] and flag[2] and not flag[3] and a[i][j] == 'a':
                flag[3] = True
                break

    if all(flag):
        print("YES")
    else:
        print("NO")

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        solve()

B.Sequence Game

a数组里，大于等于前一个值的a[i]会被写到b里。直接遍历b，如果b[i]比前一个小就在它前面插一个相等的值

def solve():
    n = int(input())
    a = []
    x_values = list(map(int, input().split()))
    a.append(x_values[0])
    
    for i in range(1, n):
        x = x_values[i]
        if x < a[-1]:
            a.append(x)
            a.append(x)
        else:
            a.append(x)
    
    print(len(a))
    print(*a)

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        solve()

C.Flower City Fence

计算反过来的长度，判断是否相等就行

def solve():
    n = int(input())
    a = list(map(int, input().split()))

    if n == 1:
        if a[0] == 1:
            print("YES")
        else:
            print("NO")
        return

    b = [0] * (n + 1)
    j = n
    for i in range(1, n + 1):
        flag = 1
        while j >= 1:
            if a[j - 1] >= i:
                b[n - i + 1] = j
                flag = 0
                break
            j -= 1
        if flag:
            print("NO")
            return

    for i in range(1, n + 1):
        if a[i - 1] != b[n - i + 1]:
            print("NO")
            return

    print("YES")

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        solve()

D.Ice Cream Balls

对于没有重复口味的集合，能选出的方案是n*(n-1)/2 （从n个里面选2个的组合数）。二分找出需要多少不同口味的冰淇淋，其余用重复口味填充

def solve():
    n = int(input())
    l, r = 2, 2648956421
    while l < r:
        mid = (l + r + 1) // 2
        if mid * mid - mid - 2 * n <= 0:
            l = mid
        else:
            r = mid - 1
    
    if l * l - l - 2 * n == 0:
        print(l)
    else:
        ans = n - l * (l - 1) // 2
        print(ans + l)

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        solve()

E.Kolya and Movie Theatre

枚举最后看哪场 Onlogn

import heapq

def solve():
    n, m, d = map(int, input().split())
    a = [0] + list(map(int, input().split()))
    f = [0] * (n + 1)
    q = []
    sum_value = 0

    for i in range(1, n + 1):
        if a[i] <= 0:
            f[i] = f[i - 1]
            continue
        if len(q) < m:
            sum_value += a[i]
            heapq.heappush(q, a[i])
        else:
            sum_value -= q[0]
            heapq.heappop(q)
            sum_value += a[i]
            heapq.heappush(q, a[i])
        f[i] = sum_value

    res = 0
    for i in range(1, n + 1):
        res = max(res, f[i] - i * d)
    
    print(res)

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        solve()

F.Magic Will Save the World（补

二分答案，注意check

#include
using namespace std;
typedef long long ll;
ll w,f,n,sum;
ll a[105];

inline bool check(ll x)
{
	ll w1=w*x,f1=f*x;
	if(w1+f1g(w1+5,0);
	int cx=0;
	g[0]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=w1;j>=a[i];j--)
		{
			g[j]=(g[j]|g[j-a[i]]);
			if(g[j])cx=max(j,cx);
		}
	}
	if(sum-cx<=f1)
		return 1;
	return 0;
}
void solve()
{
	cin>>w>>f>>n;
	if(w>a[i];
		sum+=a[i];
	}
	ll l=0,r=sum/f+10;
	while(l>1;
		if(check(mid))r=mid;
		else l=mid+1;
	}
	cout<>t;
	while(t--)solve();
	return 0;
}