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;
}