CF
Problem - 1084C - Codeforces
简单题，算每一块中a的数量，对于每一块，有选0，1，2……，k种，k+1种选法
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 1e9+7;
const int N=2e5+10;int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);string s;cin >> s;vector<int> a;for (int i = 0; i < s.size(); i++){int cnt = 0;while (i < s.size() && s[i] != 'b'){if (s[i] == 'a'){cnt++;}i++;}a.push_back(cnt);}int res = 1;for (int i = 0; i < a.size(); i++){res = ((LL)res * (a[i] + 1)) % mod;}cout << res - 1 << endl; // 删去全不选的情况
}
CF1063B
Problem - B - Codeforces
思维题，卡了两小时还没做出来
今天这场只做出来A呜呜呜
5次的操作数，就应该去考虑怎么通过5次操作把所有数覆盖住
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[N], pos[N];void solve()
{int n;cin >> n;for (int i = 0; i < n;i++){cin >> a[i];pos[a[i]] = i;}string s;cin >> s;if(s[0]=='1'||s[s.size()-1]=='1'||s[pos[1]]=='1'||s[pos[n]]=='1'){cout << -1 << endl;return;}cout << 5 << endl;cout << 1 << " " << pos[1] + 1 << endl;cout << 1 << " " << pos[n] + 1 << endl;cout << pos[1] + 1 << " " << n << endl;cout << pos[n] + 1 << " " << n << endl;cout << 1 << " " << n << endl;
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T;cin >> T;while (T--){solve();}
}