题目链接
A、Acacius and String
题意： 给你一个长度为n的字符串s，由小写英文字母和问号组成。是否可以用小写英文字母替换问号，使字符串“abacaba”在结果字符串中作为子字符串出现且仅出现一次。
思路：这题就真的暴力……暴力查找符合该串的即可……
参考了大佬的博客
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f
const int N = 2e5 + 10;
const int maxn=4000;
const double PI= acos(-1.0);
string s = "abacaba";
int findd(string str)
{
int i, j;
int cnt = 0;
for (i = 0; i <= str.size()-7; i++)
{
int flag = 1;
for (j = 0; j < 7; j++)
{
if (s[j] != str[i + j])
{
flag = 0;
break;
}
}
if (flag)
cnt++;
}
return cnt;
}
int main()
{
int t,n;
string str;
cin >> t;
while (t--)
{
cin >> n >> str;
int cnt = findd(str);
int i, j;
if (cnt > 1)
{
cout << "No" << endl;
continue;
}
if (cnt == 1)
{
cout << "Yes" << endl;
for (i = 0; i < str.size(); i++)
{
if (str[i] != '?')cout << str[i];
else cout << 'z';
}
cout << endl;
continue;
}
string tmp = str;
int flag1 = 1;
for (i = 0; i <= str.size()-7; i++)
{
int flag = 1;
str = tmp;
for (j = 0; j < 7; j++)
{
if (str[i+j] == '?') str[i + j] = s[j];
if (str[i+j] != s[j])flag = 0;
}
if (flag)
{
cnt = findd(str);
if (cnt == 1)
{
flag1 = 0;
cout << "Yes" << endl;
for (i = 0; i < str.size(); i++)
{
if (str[i] != '?')cout << str[i];
else cout << 'z';
}
cout << endl;
}
if (flag1 == 0)break;
}
}
if (flag1)
{
cout << "No" << endl;
}
}
return 0;
}B、Dubious Cyrpto
题意：有三个整数a,b,c。满足l<=a,b,c<=r 还有一个整数m=n*a+b-c。n是严格大于0的正整数。给定l,r,m。要求出a,b,c的值。
思路：枚举a即可。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f
const int N = 2e5 + 10;
const double PI= acos(-1.0);
int main()
{
int t;
cin >> t;
while(t--)
{
ll m,l,r,a,b,c;
cin >> l >> r >> m;
ll mx=r-l,mi=l-r;
for(a=l; a<=r; ++a)
{
ll x=m%a;
ll y=a-x;
if(y<=mx)
{
b=l;
c=l+y;
break;
}
else if(-x>=mi)
{
b=r;
c=r-x;
break;
}
}
cout << a << " " << b << " " << c <<endl;
}
return 0;
}