1 条题解
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1
长代码,短代码,只要能AC就是好代码
#include<iostream> #include<string> using namespace std; int main(){ string a,b; cin >> a; for(int i=0;i < a.size();i++){ if(a[i] >= 'A' && a[i] <= 'Z') a[i] += 32; } cin.ignore(); getline(cin,b); for(int i=0;i < b.size();i++){ if(b[i] >= 'A' && b[i] <= 'Z') b[i] += 32; } int ans = 0; int first = -1; size_t pos = 0; while((pos = b.find(a, pos)) != string::npos){ bool left_ok = false; bool right_ok = false; if(pos == 0 || b[pos - 1] == ' ') left_ok = true; size_t endp = pos + a.size(); if(endp == b.size() || b[endp] == ' ') right_ok = true; if(left_ok && right_ok){ if(ans == 0) first = pos; ans++; pos += a.size(); }else{ pos++; } } if(ans == 0){ cout << -1 << endl; }else{ cout << ans << " " << first << endl; } return 0; }
信息
- ID
- 5427
- 时间
- 1000ms
- 内存
- 125MiB
- 难度
- 3
- 标签
- 递交数
- 2
- 已通过
- 1
- 上传者