1132 Cut Integer(20 分)
题意:将一个含K(K为偶数)个数字的整数Z割分为A和B两部分,若Z能被A*B整除,则输出Yes,否则输出No。
分析:当A*B为0的时候,不能被Z整除,输出No。否则会出现浮点错误。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
#include<vector>
#include<set>
#include<cmath>
using namespace std;
char s[20];
int main(){int N;scanf("%d", &N);while(N--){scanf("%s", s);int len = strlen(s);int A = 0;for(int i = 0; i < len / 2; ++i){A = A * 10 + s[i] - '0';}int B = 0;for(int i = len / 2; i < len; ++i){B = B * 10 + s[i] - '0';}int C = A * B;if(C == 0){printf("No\n");continue;}int x = atoi(s);if(x % C == 0) printf("Yes\n");else printf("No\n");}return 0;
}
1133 Splitting A Linked List(25 分)
题意:给定一个链表,将链表重新排序,在不打乱原链表相对顺序的前提下,小于0的在最前面,其次是0~K,最后是大于K的数。
分析:
1、3次遍历可实现链表重排。
2、map映射value和pre或suc的关系会超时,所以,以pre为结点定义结构体,组织链表的重排,从而进行优化。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
#include<vector>
#include<set>
#include<cmath>
using namespace std;
const int MAXN = 100000 + 10;
struct Node{int pre, value, suc;
}num[MAXN];
vector<int> old, ans;
int main(){int N, K, head, pre, value, suc;scanf("%d%d%d", &head, &N, &K);for(int i = 0; i < N; ++i){scanf("%d%d%d", &pre, &value, &suc);num[pre].pre = pre;num[pre].value = value;num[pre].suc = suc;}while(head != -1){old.push_back(head);head = num[head].suc;}int len = old.size();for(int i = 0; i < len; ++i){if(num[old[i]].value < 0) ans.push_back(old[i]);}for(int i = 0; i < len; ++i){if(num[old[i]].value >= 0 && num[old[i]].value <= K) ans.push_back(old[i]);}for(int i = 0; i < len; ++i){if(num[old[i]].value > K) ans.push_back(old[i]);}for(int i = 0; i < len - 1; ++i){printf("%05d %d %05d\n", ans[i], num[ans[i]].value, ans[i + 1]);}printf("%05d %d -1\n", ans[len - 1], num[ans[len - 1]].value);return 0;
}
1134 Vertex Cover(25 分)
题意:vertex cover是指图中一些点的集合,使得图中每一条边的两个点中都至少有一个点在该点集中。给定点的集合,判断是否为vertex cover。
分析:按题意模拟即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
#include<vector>
#include<set>
#include<cmath>
using namespace std;
const int MAXN = 10000 + 10;
int N, M;
bool vis[MAXN];
struct Edge{int u, v;void read(){scanf("%d%d", &u, &v);}
}num[MAXN];
int main(){scanf("%d%d", &N, &M);for(int i = 0; i < M; ++i){num[i].read();}int K;scanf("%d", &K);while(K--){int n, x;scanf("%d", &n);memset(vis, false, sizeof vis);while(n--){scanf("%d", &x);vis[x] = true;}bool ok = true;for(int i = 0; i < M; ++i){if(vis[num[i].u] || vis[num[i].v]) continue;ok = false;break;}if(ok) printf("Yes\n");else printf("No\n");}return 0;
}
1135 Is It A Red-Black Tree(30 分
题意:给定一棵二叉搜索树的前序遍历序列,判断其是否为一棵红黑树。
分析:
1、红黑树是一棵平衡的二叉搜索树,满足以下条件:
(1)所有结点不是红色就是黑色;
(2)根结点是黑色;
(3)每一个为NULL的叶子结点是黑色;
(4)如果某结点是红色,其左右子结点都是黑色;
(5)对于每个结点,其到所有后代叶子结点经过的黑色结点数相同;
2、根据给定的前序遍历序列,结合二叉搜索树的定义可以建树。
3、递归检查条件4。
4、同理,递归统计左右子树的黑色结点数,来检查条件5。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
#include<vector>
#include<set>
#include<cmath>
using namespace std;
const int MAXN = 30 + 10;
struct Node{Node *left, *right;int value;
};
struct Node *root;
bool ok;
void build(Node* &r, int x){if(r == NULL){r = (Node*)malloc(sizeof(Node));r -> value = x;r -> left = r -> right = NULL;return;}if(abs(x) < abs(r -> value)){build(r -> left, x);}else{build(r -> right, x);}
}
void judge_RedNode(Node* r){if(!ok) return;if(r -> left != NULL){if(r -> value < 0 && r -> left -> value < 0){ok = false;return;}else judge_RedNode(r -> left);}if(r -> right != NULL){if(r -> value < 0 && r -> right -> value < 0){ok = false;return;}else judge_RedNode(r -> right);}
}
int judge_BlackNode(Node* r){int leftcnt, rightcnt;if(!ok) return -1;if(r == NULL) return 1;leftcnt = judge_BlackNode(r -> left);rightcnt = judge_BlackNode(r -> right);if(leftcnt != rightcnt){ok = false;return -1;}else{if(r -> value > 0) ++leftcnt;}return leftcnt;
}
int main(){int K;scanf("%d", &K);while(K--){root = NULL;int N, x;scanf("%d", &N);while(N--){scanf("%d", &x);build(root, x);}ok = true;judge_RedNode(root);judge_BlackNode(root);if(root -> value < 0 || !ok){printf("No\n");}else{printf("Yes\n");}}return 0;
}