PAT 天梯赛真题集
题目：L2-010 排座位
题意：
1. x与y是敌对关系： a)也有共同好友：OK but...
b)无共同朋友：No way
2. x与y是朋友关系：No problem
3. x与y既不是朋友也不敌对：OK
朋友间并查集，用map记录敌对关系。
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#include <bits/stdc++.h>#define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define scf2(a,b) scanf("%d%d",&a,&b)using namespace std;int fa[110];int find(int x){return fa[x] == x ? x : fa[x] = find(fa[x]);}int main(){ map< pair<int,int> ,bool>mp; int n, m, k; scf3(n,m,k); int x, y , op; for(int i = 1; i <= n; i++) fa[i] = i; for(int i = 1; i <= m; i++){ scf3(x,y,op); if(op == 1){ fa[find(x)] = find(y); } else if(op == -1){ mp[make_pair(x,y)] = false; mp[make_pair(y,x)] = false; } } while(k--){ scf2(x,y); if(find(x) == find(y)){ if( !mp.count(make_pair(x,y)) || !mp.count(make_pair(y,x)) ) //不敌对 puts("No problem"); else puts("OK but..."); } else{ if( !mp.count(make_pair(x,y)) || !mp.count(make_pair(y,x)) ) //不敌对 puts("OK"); else puts("No way"); } }} |
题意：模拟，sort.
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#include <bits/stdc++.h>using namespace std;struct Node{ int num; int get; double val;}node[10010];bool cmp(const Node &a,const Node &b){ //核心排序代码 if(a.val == b.val) { if(a.get == b.get) return a.num < b.num; else return a.get > b.get; } else return a.val > b.val;}int main(){ int n, m; memset(node,0,sizeof(node)); scanf("%d",&n); for(int i = 1; i <= n; i++) //初始化 下标等于编号 node[i].num = i; for(int i = 1; i <= n; i++){ scanf("%d",&m); int x; double y; while(m--){ scanf("%d%lf", &x, &y); node[x].val += y; node[x].get ++; node[i].val -= y; } } sort(node + 1, node + n + 1, cmp); for(int i = 1; i <= n; i++){ printf("%d %.2f\n",node[i].num,node[i].val/100); }} |
题目：L2-008 最长对称子串 用力戳我直达原题
题意：暴力，O(n2)
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#include <bits/stdc++.h>using namespace std;int main(){ string s; getline(cin,s); int ans = -1; for(int i = 0; i < s.size(); i++){ // aba型 int sum = 1; int idx = i + 1; for(int j = i - 1; j >= 0; j--){ if(s[idx] != s[j]) break; else sum += 2; idx++; if(idx >= s.size()) break; } ans = max(ans,sum); } for(int i = 0; i < s.size(); i++){ //aabb型 if(s[i] != s[i+1]) continue; int sum = 2; int idx = i + 2; for(int j = i - 1; j >= 0; j--){ if(s[idx] != s[j]) break; else sum += 2; idx++; if(idx >= s.size()) break; } ans = max(ans,sum); } cout << ans << endl;} |
题目：L2-016. 愿天下有情人都是失散多年的兄妹
用力戳我直达原题题意：就是....不能近亲结婚...也不能同性恋....
用二叉树存双亲，DFS搜x的前四代人，存进set，再搜y的前四代人，看是否存在于set
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#include <bits/stdc++.h>using namespace std;set<int>st;bool flag;struct Node{ int fa,mon; char sex;}node[100100];void dfs(int x,int num){ if(num > 5) return; st.insert(x); if(node[x].fa != -1) find(node[x].fa,num + 1); if(node[x].mon != -1) find(node[x].mon,num + 1);}void judge(int x,int num){ if(num > 5) return; if(st.find(x) != st.end()){ flag = false; return; } if(node[x].fa != -1) judge(node[x].fa, num + 1); if(node[x].mon != -1) judge(node[x].mon,num + 1);}int main(){ int n, id, fa, mon; char sex; memset(node,-1,sizeof(node)); scanf("%d",&n); while(n--){ scanf("%d %c %d %d",&id, &sex, &fa, &mon); node[id].fa = fa; node[id].mon = mon; node[id].sex = sex; node[fa].sex = 'M'; //父母也要设置性别啊...不然扣了8分orz node[mon].sex = 'F'; } scanf("%d",&n); int x, y; while(n--){ cin >> x >> y; if(node[x].sex == node[y].sex){ puts("Never Mind"); continue; } st.clear(); flag = true; dfs(x,1); //深搜将四代存进set judge(y,1); //深搜看四代是否与x重，用st.find() if(flag == true) puts("Yes"); else puts("No"); }} |
题目：L2-015. 互评成绩 用力戳我直达原题
题意： 模拟，sort，最后存stack。如此水题，去年决赛时竟没做满分，而且弄得很复杂.......
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#include <bits/stdc++.h>using namespace std;bool cmp(const int &a, const int &b){return a > b;}int main(){ int n, k , m, x; double sco[10010]; scanf("%d%d%d",&n, &k, &m); for(int i = 0; i < n; i++){ int maxx = -1, minn = 110; for(int j = 0; j < k; j++){ scanf("%d",&x); maxx = max(maxx, x); minn = min(minn, x); sco[i] += x; } sco[i] -= (maxx + minn); //去掉最高分和最低分 } sort(sco, sco + n, cmp); stack<double>sta; for(int i = 0; i < m; i++){ sta.push(sco[i] / (k - 2)); } bool cnt = false; while(!sta.empty()){ //倒序输出 if(cnt) printf(" "); cnt = true; printf("%.3f",sta.top()); sta.pop(); }} |
题目：L1-006 抢红包 用力戳我直达原题
题意：求最长的 最小连续因子
做法：三个for暴力枚举。注意len==1也要跑，因为例如 9 * 9，结果是1 3 ，而不是 1 9
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#include <bits/stdc++.h>#define scf0(a) scanf("%s",&a)#define scf1(a) scanf("%d",&a)#define scf2(a,b) scanf("%d%d",&a,&b)#define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define MEM(a,b) memset(a,b,sizeof(a))#define pii pair<int,int>#define pdd pair<double,double>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int maxn = 10000 + 5;int main() { int n, len, i, j; scf1(n); bool flag = false; for(len = 12; len >= 1; len--) { //长度从12往1搜 for(i = 2; i <= (int)sqrt(n); i++) { //i代表连续因子的第一个数，因子最大只能是sqrt(n) LL sum = 1; for(j = i; j <= len+i-1; j++) { //连续len个数相乘 sum *= j; } if(n % sum == 0) { flag = true; break; } } if(flag) break; } if(flag) { printf("%d\n",len); bool cnt = false; for(int k = i; k <= len+i-1; k++) { if(cnt) printf("*"); cnt = true; printf("%d",k); } puts(""); } else { printf("1\n%d",n); }} |
