C. Unstable String
题意:读来读去,意思就是给你一个字符串,1010101和010101 这种形式的字符串是美丽的,?可以变成0或1,求所给字符串有多少子串是美丽的。
思路:
题解
考虑
d
p
dp
dp
f
[
i
]
[
j
]
f[i][j]
f[i][j] 表示第
i
i
i 位为
j
j
j,以
i
i
i 结尾的不稳定子串数量
注意,我们可以对
?
?
? 分别取
0
0
0 和
1
1
1 计两次贡献
code:
#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e6 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
int f[maxn][2];
string s;
void work()
{
cin >> s;n = s.size();s = "@" + s;
ll ans = 0;
for(int i = 1; i <= n; ++i) f[i][0] = f[i][1] = 0;
for(int i = 1; i <= n; ++i){
if(s[i] == '?'){
f[i][0] = f[i-1][1] + 1; f[i][1] = f[i-1][0] + 1;
}
else if(s[i] == '0') f[i][0] = f[i-1][1] + 1;
else f[i][1] = f[i-1][0] + 1;
ans += max(f[i][0], f[i][1]);
}
cout << ans << endl;
}
int main()
{
ios::sync_with_stdio(0);
int TT;cin>>TT;while(TT--)
work();
return 0;
}
D. Playoff Tournament
思路:
题解
每次询问整个区间内可能成为胜者的数量
把编号倒过来就变成线段树结构了
每次查询修改叶子节点,完事递归向上更新信息即可即可
复杂度
O
(
n
+
q
k
)
O(n+qk)
O(n+qk)
code:
#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e6 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
string s = "@", t;
int ans[maxn];
#define ls p << 1
#define rs p << 1 | 1
void build(int p = 1, int cl = 1, int cr = n){
if(cl == cr){
ans[p] = 1;return;
}
int mid = cl + cr >> 1;
build(ls, cl, mid); build(rs, mid + 1, cr);
if(s[p] == '0') ans[p] = ans[rs];// 向上合并更新信息
else if(s[p] == '1') ans[p] = ans[ls];
else ans[p] = ans[ls] + ans[rs];
}
void update(int p){// 向上合并更新信息
if(!p) return;
if(s[p] == '0') ans[p] = ans[rs];
else if(s[p] == '1') ans[p] = ans[ls];
else ans[p] = ans[ls] + ans[rs];
update(p >> 1);
}
void work()
{
cin >> m >> t;
n = 1 << m;t = "@" + t;
for(int i = 1; i < n; ++i) s.push_back(t[n - i]);// 翻转字符串就可以得到新的编号
int q;cin >> q;
build();
while(q--){
int x;char y;cin >> x >> y;
x = n - x;s[x] = y;
update(x);
cout << ans[1] << endl;
}
}
int main()
{
ios::sync_with_stdio(0);
// int TT;cin>>TT;while(TT--)
work();
return 0;
}