PAT乙级题目题解
1003. 我要通过！(20)
思路：
仔细读题....其实代码很简单
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 bool judge(char s[]) 10 { 11 int len = strlen(s); 12 int cnt_p = 0, cnt_a = 0, cnt_t = 0; 13 for(int i = 0; i < len; i++) 14 { 15 cnt_p += s[i] == 'P'; 16 cnt_a += s[i] == 'A'; 17 cnt_t += s[i] == 'T'; 18 } 19 20 if(cnt_p != 1 || cnt_t != 1 || !cnt_a || cnt_p + cnt_a + cnt_t != len) return false; 21 22 int x = strchr(s, 'P') - s; 23 int y = strchr(s, 'T') - strchr(s, 'P') - 1; 24 int z = len - 1 - (strchr(s, 'T') - s); 25 return x * y == z; 26 } 27 28 int main() 29 { 30 int n; 31 char s[105]; 32 scanf("%d", &n); 33 while(n--) 34 { 35 scanf("%s", s); 36 if(judge(s)) puts("YES"); 37 else puts("NO"); 38 39 } 40 return 0; 41 }
1005. 继续(3n+1)猜想 (25)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int a[105]; 10 11 int main() 12 { 13 set<int>s; 14 int n; 15 scanf("%d", &n); 16 for(int i = 0; i < n; i++) 17 { 18 scanf("%d", &a[i]); 19 s.insert(a[i]); 20 } 21 for(auto o : s) 22 { 23 while(o != 1) 24 { 25 if(o & 1) o = (3 * o + 1) / 2; 26 else o /= 2; 27 if(s.find(o) != s.end()) s.erase(o); 28 } 29 } 30 31 set<int>::reverse_iterator it = s.rbegin(); cout << *it++; 32 while(it != s.rend()) cout << " " << *it++; 33 34 puts(""); 35 return 0; 36 }
1007. 素数对猜想 (20)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int prime[9600], check[100000 + 5]; 10 int sieve(LL n) 11 {//O(n)复杂度的线性筛。 接受不了的可以先看埃氏筛法 12 memset(check, false, sizeof(check)); 13 check[0] = check[1] = true; 14 int tot = 0; 15 for(LL i = 2; i <= n; i++) 16 { 17 if(!check[i]) prime[tot++] = i; 18 for(LL j = 0; j < tot && i * prime[j] <= n; j++) 19 { 20 check[i * prime[j]] = true; 21 if(i % prime[j] == 0) break; 22 } 23 } 24 return tot; 25 } 26 27 28 int main() 29 { 30 int n; 31 cin >> n; 32 int tot = sieve(n); 33 int ans = 0; 34 for(int i = 1; i < tot; i++) 35 { 36 if(prime[i] - prime[i - 1] == 2) ans++; 37 } 38 cout << ans << endl; 39 return 0; 40 }
1008. 数组元素循环右移问题 (20)
思路：
倍增这个数组，直接输出就好了。另外m可能远大于n，所以先m %= n;
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int a[205]; 10 11 int main() 12 { 13 int n, m; 14 scanf("%d%d", &n, &m); 15 m %= n;//m可能远大于n 16 for(int i = 0; i < n; i++) 17 { 18 scanf("%d", &a[i]); 19 a[i + n] = a[i]; 20 } 21 22 for(int i = n - m; i <= n + n - m - 1; i++) 23 { 24 printf("%d%c", a[i], i == (2 * n - m - 1) ? '\n' : ' '); 25 } 26 return 0; 27 }
1009. 说反话 (20)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 10 int main() 11 { 12 stack<string>sta; 13 string s; 14 while(cin >> s) 15 { 16 sta.push(s); 17 char ch = getchar(); 18 if(ch == '\n') break; 19 } 20 while(sta.size()) 21 { 22 cout << sta.top(); 23 sta.pop(); 24 if(sta.size()) cout << " "; 25 else cout << endl; 26 } 27 return 0; 28 }
1010. 一元多项式求导 (25)
思路：
这题坑点比较多。。要考虑b==0 b!=0 b==0且只有一组数之类的情况...比较恶心
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 10 int main() 11 { 12 char ch; 13 int a, b; 14 bool first = true; 15 while(~scanf("%d %d%c", &a, &b, &ch)) 16 { 17 if(b != 0) 18 { 19 if(first) 20 printf("%d %d", a * b, b - 1); 21 else 22 printf(" %d %d", a * b, b - 1); 23 first = false; 24 } 25 26 if(ch == '\n') 27 { 28 if(first && b == 0) puts("0 0"); 29 else puts(""); 30 break; 31 } 32 } 33 return 0; 34 }
1012. 数字分类 (20)
思路：
纯模拟没啥好说的.... 就是注意判空的情况就好了。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 vector<int>vec[6]; 9 int main() 10 { 11 int n; 12 cin >> n; 13 for(int i = 0; i < n; i++) 14 { 15 int x; 16 scanf("%d", &x); 17 vec[x % 5].push_back(x); 18 } 19 20 int a1 = 0, ok = 0; 21 for(auto o : vec[0]) if(o % 2 == 0) a1 += o, ok = 1; 22 if(ok) printf("%d ", a1); 23 else printf("N "); 24 25 int a2 = 0; 26 for(int i = 0; i < vec[1].size(); i++) 27 { 28 if(i & 1) a2 -= vec[1][i]; 29 else a2 += vec[1][i]; 30 } 31 if(vec[1].size()) printf("%d ", a2); 32 else printf("N "); 33 34 int a3 = vec[2].size(); 35 if(vec[2].size()) printf("%d ", a3); 36 else printf("N "); 37 38 double a4; 39 if(vec[3].size()) 40 { 41 int sum = 0; 42 for(auto o : vec[3]) sum += o; 43 a4 = 1.0 * sum / vec[3].size(); 44 printf("%.1f ", a4); 45 } 46 else printf("N "); 47 48 49 if(vec[4].size()) 50 { 51 int a5 = *max_element(vec[4].begin(), vec[4].end()); 52 printf("%d\n", a5); 53 } 54 else printf("N\n"); 55 return 0; 56 }
1014. 福尔摩斯的约会 (20)
题目：
这题很没意思.....就是题目给的字符串不一定满足条件，你要自己控制一下两个字符相等的时候，满不满足时间/星期的要求，会不会越过去，比如星期的时候，字母<='G'之类的orz。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 10000 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 string day[] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"}; 9 int main() 10 { 11 string a, b, c, d; 12 cin >> a >> b >> c >> d; 13 int len = min(a.length(), b.length()); 14 int cnt = 1; 15 for(int i = 0; i < len; i++) 16 { 17 if(a[i] == b[i]) 18 { 19 if(cnt == 1) 20 { 21 if(a[i] >= 'A' && a[i] <= 'G') cout << day[a[i] - 'A'] << " "; 22 else continue; 23 } 24 else if(cnt == 2) 25 { 26 int temp = 0; 27 if(a[i] >= 'A' && a[i] <= 'N') temp = a[i] - 'A' + 10; 28 else if(a[i] >= '0' && a[i] <= '9') temp = a[i] - '0'; 29 else continue; 30 printf("%02d:", temp); 31 break; 32 } 33 cnt ++; 34 } 35 } 36 len = min(c.length(), d.length()); 37 for(int i = 0; i < len; i++) 38 { 39 if(c[i] == d[i] && isalpha(c[i])) 40 { 41 printf("%02d\n", i); 42 break; 43 } 44 } 45 return 0; 46 }
1015. 德才论 (25)
思路：
用一个level来记录分类，会方便一些。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 1e5 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 struct node 10 { 11 int num, de, cai, sum, level; 12 bool operator < (const node &other)const 13 { 14 if(level != other.level) 15 return level < other.level; 16 if(sum != other.sum) 17 return sum > other.sum; 18 if(de != other.de) 19 return de > other.de; 20 if(num != other.num) 21 return num < other.num; 22 } 23 }nodes[maxn]; 24 25 int main() 26 { 27 int n, low, high; 28 scanf("%d%d%d", &n, &low, &high); 29 int cnt = 0; 30 for(int i = 0; i < n; i++) 31 { 32 int num, de, cai, level; 33 scanf("%d%d%d", &num, &de, &cai); 34 35 if(de < low || cai < low) level = 5, cnt++; 36 else if(de >= high && cai >= high) level=1; 37 else if(de >= high && cai < high) level=2; 38 else if(de < high && cai < high && de >= cai) level=3; 39 else if(de >= low && cai >= low) level=4; 40 41 nodes[i] = {num, de, cai, de + cai, level}; 42 } 43 sort(nodes, nodes + n); 44 printf("%d\n", n - cnt); 45 for(int i = 0; i < n - cnt; i++) 46 printf("%d %d %d\n", nodes[i].num, nodes[i].de, nodes[i].cai); 47 48 return 0; 49 }
1017. A除以B (20)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 1e5 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int main() 10 { 11 string a; 12 int b; 13 cin >> a >> b; 14 int len = a.length(); 15 int temp = a[0] - '0'; 16 if(len == 1) 17 { 18 printf("0 %d\n", temp); 19 return 0; 20 } 21 22 for(int i = 1; i < len; i++) 23 { 24 temp = temp * 10 + a[i] - '0'; 25 printf("%d", temp / b), temp %= b; 26 } 27 printf(" %d", temp); 28 return 0; 29 }
1019. 数字黑洞 (20)
思路：
做这种题就小心一点就行了，样例读出来0000也能终止，然后看一下输入要求1-10000内正整数，就意味着可能456,23这种输入也有。所以求余要小心点。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 1e5 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int main() 10 { 11 int n, d; 12 cin >> n; 13 d = n; 14 do 15 { 16 int cnt = 0, a[5]; 17 for(int i = 0; i < 4; i++) 18 { 19 a[i] = d % 10, d /= 10; 20 } 21 sort(a, a + 4); 22 int x = 0, y = 0; 23 for(int i = 0; i < 4; i++) x = x * 10 + a[i]; 24 for(int i = 3; i >=0; i--) y = y * 10 + a[i]; 25 d = y - x; 26 printf("%04d - %04d = %04d\n", y, x, d); 27 }while(d != 6174 && d != 0); 28 return 0; 29 }
1021. 个位数统计 (15)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 1e5 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int main() 10 { 11 string s; 12 cin >> s; 13 int len = s.length(); 14 map<int, int>ma; 15 for(int i = 0; i < len; i++) 16 { 17 ma[s[i] - '0'] ++; 18 } 19 for(int i = 0; i < 10; i++) 20 { 21 if(ma.count(i) != 0) 22 { 23 printf("%d:%d\n", i, ma[i]); 24 } 25 } 26 return 0; 27 }
1022. D进制的A+B (20)
思路：
坑点：进制转换的时候一定要小心0的转换要特判。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int mod = 1e9 + 7; 4 const int maxn = 1e5 + 5; 5 const int INF = 0x3f3f3f3f; 6 typedef long long LL; 7 typedef unsigned long long ull; 8 9 int main() 10 { 11 int a, b, d; 12 cin >> a >> b >> d; 13 a += b; 14 //(a)10 -> (x)d 15 stack<int>s; 16 if(a == 0) 17 { 18 puts("0"); 19 return 0; 20 } 21 while(a) 22 { 23 s.push(a % d); 24 a /= d; 25 } 26 while(s.size()) printf("%d", s.top()), s.pop(); 27 puts(""); 28 return 0; 29 }
1023. 组个最小数 (20)
