1、最小子串覆盖
考点:双指针
代码:
def judge(s,t,l,r):
m = [0] * (256)
for i in t:
m[ord(i)] += 1
for i in range(l,r+1):
if s[i] in t:
m[ord(s[i])] -= 1
for i in m:
if i > 0:
return False
return True
def main():
result = "No"
s = input()
t = input()
l = r = 0 #定义一个滑动窗口
num = 100000
# print("字符串",len(s))
while(r < len(s) and l<=r):
if judge(s,t,l,r):
# tmp = s[l:r + 1]
# print("满足条件的字符串:",tmp)
l += 1
while judge(s,t,l,r):
# tmp = s[l:r + 1]
# print("滑动窗口缩小:",tmp)
l+=1
l -= 1
if (r-l+1) < num:
num = r-l+1
result = s[l:r + 1]
r += 1
print(result)
if __name__ == '__main__':
main();
2.方块桶
代码:
def main():
n = int(input())
a = [int(i) for i in input().split()]
# print(n,a)
l = 0
r = n - 1
maxl = a[0]
maxr = a[n-1]
ans = 0
while l
3.新材料
#include
using namespace std;
map lst;
int n,k;
int main( )
{
scanf("%d%d",&n,&k);
int ans=0;
for(int i=1;i<=n;i++){
int cur;
scanf("%d",&cur);
if(!lst.count(cur)) lst[cur] = i;
else if (lst[cur] == -1) continue;
else if(i - lst[cur] > k) lst[cur] = i;
else ans ^= cur,lst[cur] = -1;
}
cout << ans << endl;
}
4、与蝴蝶一起消散吧
解题思路:看不懂看不懂
直接抄答案
#include
using namespace std;
int n,k;
int main( )
{
long long ans0 = 0,ans1 = 0;
scanf("%d%d",&n,&k);
for(int i = 1;i <= n;i++){
int a,m;
scanf("%d%d",&a,&m);
if(a>k){
ans0 = ans1 = min(ans0 + k,ans1) + 1ll * (a-k) *m;
}
else{
long long nxt0 = min(ans0 + 1ll * (m + 1)/2*a,ans1 + 1ll*m/2*a);
long long nxt1 = min(ans0 + 1ll * (m + 2)/2*a,ans1 + 1ll*(m + 1)/2*a);
ans0 = nxt0;
ans1 = nxt1;
}
}
cout << min(ans0,ans1)<
5、炼金术
这班实在是呆不下去了,这个题也太长了,答案都那么长,抄都不想抄
给大家一个链接吧,自己抄吧
【2023百度之星第三场题解】嘉宾:NOI、IOI金牌周航锐_哔哩哔哩_bilibili