PTA 7-17（查找） 字符串关键字的散列映射（25 分） 25分代码

需要注意的就是冲突的处理那部分，数据结构课本上的知识，运用 平方探测法解决

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<map>

using namespace std;
const int maxn = 1000000 + 7, INF = 0x7f7f7f7f, mod = 1e9+7;
int n, m;
string s;// t;
map<string, int> mp;
bool vis[maxn] = {false};

int main() {
scanf("%d %d", &n, &m);
for(int i = 1; i <= n; ++i) {
cin >> s; int len = s.size();
int num = 0;
if(len == 1) {
num += (s[0]-'A');
}
else if(len == 2) {
num = 32*(s[0]-'A') + (s[1]-'A');
}
else {
for(int j = 3; j >= 1; --j) {
int pos = len - j;
num = num * 32 + s[pos] - 'A';
}
}

num %= m;

if(mp[s] == 0 && vis[num]) {
for(int t = 1; t < maxn; ++t) {
if(!vis[(num+t*t)%m]) {
num = (num+t*t)%m;
vis[num] = true;
mp[s] = num;
cout << num;
break;
}
else if(!vis[(num-t*t+m)%m]) {
num = (num-t*t+m)%m;
vis[num] = true;
mp[s] = num;
cout << num;
break;
}
}
}
else {
num %= m;
mp[s] = num;
cout << num;
vis[num] = true;
}

if(i < n) cout << " ";
else cout << endl;
}
return 0;
}