2024 (ICPC) Jiangxi Provincial Contest(VP补题记录)

2024 (ICPC) Jiangxi Provincial Contest(VP补题记录)

已ac 8/12，赛时7题，赛后1题。

A(签到中的签到，pass)

C(简单思维)

要么 n 要么 n-1.

FastIO 已省略

def solve():
    n,s = MI()
    a = LI()
    print1(n if sum(a) == s else n-1)
    
# for _ in range(I()):
solve()

G

给你一个长度不超过 $10^{14}$ 的11进制数，判断是不是5的倍数，由于 $11^i\mathrm{mod}10=1$，所以所有位累加即可。

mod = 5
dic = {
   
   str(i):i for i in range(10)}
dic['A'] = 10
def solve():
    n = I()
    ans = 0
    for i in range(n):
        x,y = MS()
        ans += int(x) * dic[y]
        ans %= 5
    print('Yes' if ans % 5 == 0 else 'No')

for _ in range(I()):
    solve()

# print(set([pow(11,i,10) for i in range(100000)]))

J(按题意模拟即可)

se = set()
for c in "psm":
    se.add('1'+c)
    se.add('9'+c)
for i in range(1,8):
    se.add(str(i) + 'z')
def solve():
    s = S()
    dic = {
   
   }
    for i in range(0,28,2):
        ts = s[i:i+2]
        dic[ts] = dic.get(ts,0)+1
    flag = 1
    for v in dic.values():
        if v != 2:
            flag = 0
            break
    if flag:
        print1("7 Pairs")
        return
    if se == set(dic.keys()): print1("Thirteen Orphans")
    else: print1("Otherwise")

for _ in range(I()):
    solve()

K

显然有m-1种分叉情况，所以 $ans=2^{m-1}$ % $mod$ .

mod = 998244353
def solve():
    m = I()
    print1(pow(2, m-1, mod))

solve()

H(卷积加权和反过来看)

考虑把卷积核 $K$ 加权到输入矩阵