今天下午的PAT考试状态不理想，回来怒刷了一遍，解题报告如下：
1073. Scientific Notation (20)
基本模拟题，将一长串的科学计数转换为普通的数字表示方式。思路是是数组存储输入，然后找到指数位置，并根据指数的大小和正负对前面的小数进行针对性的处理。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
class PAT{
public:
enum { N = 10000, };
char num[N];
int finger,epos,length;
void run();
};
void PAT::run()
{
scanf("%s", &num);
length = strlen(num);
epos = find(num, num + length, 'E') - num;
sscanf(num + epos + 2, "%d", &finger);
if (num[epos + 1] == '-') finger = -1 * finger;
if(num[0]=='-')printf("%c", num[0]);
int i, j;
if (finger >= 0)
{
printf("%c", num[1]);
for (i = 3; i < epos && finger > 0; i++,finger--) printf("%c", num[i]);
if (i < epos) { printf("."); for (; i < epos; i++)printf("%c", num[i]); }
else if (finger > 0) while (finger > 0){ printf("0"); finger--; }
}
else
{
printf("0."); finger++;
while (finger < 0) {printf("0"); finger++;}
printf("%c", num[1]);
for (int i = 3; i < epos; i++)printf("%c", num[i]);
}
}
int main()
{
//freopen("input.txt", "r", stdin);
PAT *p = new PAT;
p->run();
return 0;
}
PAT 1074. Reversing Linked List (25)
这个题不难，关键是要细心，首先要看懂题意。题目说的逆转排序是指每隔K个节点就将这k个节点的顺序颠倒过来，而如果最后不足K个节点的话，是不进行颠倒的，我下午考试的时候把这个意思弄错了，以为最后的节点也要进行颠倒，结果当然不对。这个题可以用map来做，不过据我同学说，用map查询会超时，我没有这样试过，不知道究竟如何，我使用的是hash的办法，直接在数组里面存储，速度会快点。
另外，这个题跟前面1052题的链表排序一样，有一个陷阱，就是不是所有给出的节点都属于此链表，需要判断，最后一个测试点是这个陷阱，不过这个点只有1分，姥姥出题的时候还是很人道。不过很多人这个点都没过。
#include <iostream>
using namespace std;
class PAT
{
public:
enum{N=100000};
struct Node
{
int addr, data, next;
};
Node node[N];
int hash[N];//addr->index
int addr[N];//index->addr
int n,k,head;
void run();
};
void PAT::run()
{
scanf("%d%d%d", &head, &n, &k);
for (int i = 0; i < n; i++)
{
scanf("%d%d%d", &node[i].addr, &node[i].data, &node[i].next);
hash[node[i].addr] = i;
}
int cur = head,c=0;
while (cur != -1)
{
addr[c] = node[hash[cur]].addr;
cur = node[hash[cur]].next;
c++; //节点计数
}
n = c; //排除不在链表中的节点
for (int i = 0; i < n; i = i + k)
{
int r = (i + k<= n) ? i + k - 1 : n - 1;
if (i + k <= n)
{
for (int j = r; j > i; j--)
{
node[hash[addr[j]]].next = node[hash[addr[j - 1]]].addr;
}
if (r == n - 1) node[hash[addr[i]]].next = -1;
else node[hash[addr[i]]].next = (i + 2 * k <= n) ? node[hash[addr[i + 2*k -1]]].addr : node[hash[addr[i+k]]].addr;
if (r!=n-1)
for (int j = r; j >= i; j--) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
else
{
for (int j = r; j > i; j--) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
printf("%05d %d -1\n", node[hash[addr[i]]].addr, node[hash[addr[i]]].data);
}
}
else
{
for (int j = i; j < n-1; j++) printf("%05d %d %05d\n", node[hash[addr[j]]].addr, node[hash[addr[j]]].data, node[hash[addr[j]]].next);
printf("%05d %d -1\n", node[hash[addr[n-1]]].addr, node[hash[addr[n-1]]].data);
}
}
}
int main()
{
//freopen("input.txt", "r", stdin);
PAT *p = new PAT;
p->run();
return 0;
}
这个题是一个常规的排序题，排序的时候仔细点。有一个要注意的地方是，如果提交了没有编译通过，submit里面显示的分数是-1，但是最终输出要输出为0。
#include <iostream>
#include <algorithm>
using namespace std;
struct User
{
int id, total, score[6], rank, solved;
bool onlist;
};
class PAT
{
public:
enum{N=10000};
User user[N];
int fullscore[N];
int n, k, m;
void run();
void printscore(const User &u);
};
bool cmp(const User &u1, const User &u2)
{
if (u1.total != u2.total) return u1.total > u2.total;
else if (u1.solved != u2.solved) return u1.solved > u2.solved;
else return u1.id < u2.id;
}
void PAT::printscore(const User &u)
{
for (int i = 1; i <= k; i++)
{
if (u.score[i] == -2) printf(" -");
else if (u.score[i] == -1) printf(" 0");
else printf(" %d", u.score[i]);
}
}
void PAT::run()
{
scanf("%d%d%d", &n, &k, &m);
for (int i = 0; i<n; i++)
{
user[i].id = i+1;
user[i].total = 0;
user[i].solved = 0;
user[i].onlist = false;
for (int j = 1; j <= k; j++)
user[i].score[j] = -2;
}
for (int i = 1; i <= k; i++) scanf("%d", &fullscore[i]);
int tid, tpid, tscore;
while (m-- > 0)
{
scanf("%d%d%d", &tid, &tpid, &tscore);
if (user[tid-1].score[tpid] < tscore) user[tid-1].score[tpid] = tscore;
}
for (int i = 0; i < n; i++)
{
for (int j = 1; j <= k; j++)
{
if (user[i].score[j] >= 0)
{
user[i].total += user[i].score[j]; user[i].onlist = true;
}
if (user[i].score[j] == fullscore[j]) user[i].solved++;
}
}
sort(user, user + n, cmp);
user[0].rank = 1;
for (int i = 1; i < n; i++)
{
if (user[i].total == user[i - 1].total) user[i].rank = user[i - 1].rank;
else user[i].rank = i+1;
}
for (int i = 0; i < n; i++)
{
if (user[i].onlist)
{
printf("%d %05d %d", user[i].rank, user[i].id, user[i].total);
printscore(user[i]);
printf("\n");
}
else break;
}
}
int main()
{
//freopen("input.txt", "r", stdin);
PAT *p = new PAT;
p->run();
return 0;
}
这个题直接用BFS搜索遍历即可。
#include <iostream>
#include <vector>
using namespace std;
class PAT
{
public:
enum{N=1001};
vector<int> fans[N];
vector<int> curfans;
int vis[N];
int n, level, k;
void run();
void bfs(int);
int maxforward;
};
void PAT::run()
{
scanf("%d%d", &n, &level);
int tnum,tfol;
for (int i = 1; i <= n; i++)
{
scanf("%d", &tnum);
while (tnum-- > 0)
{
scanf("%d", &tfol);
fans[tfol].push_back(i);
}
}
scanf("%d", &k);
int qnum;
while (k-- > 0)
{
scanf("%d", &qnum);
fill(vis, vis + N, 0);
curfans.clear();
maxforward = 0;
vis[qnum] = 1;
int fansize = fans[qnum].size();
for (int i = 0; i < fansize; i++)
{
curfans.push_back(fans[qnum][i]);
vis[fans[qnum][i]] = 1;
}
bfs(level);
printf("%d\n", maxforward);
}
}
void PAT::bfs(int l)
{
if (l == 0) return;
if (curfans.empty()) return;
int fansize = curfans.size();
maxforward += fansize;
vector<int> tmp;
for (int i = 0; i < fansize; i++)
{
int fansize2 = fans[curfans[i]].size();
vector<int> &cur = fans[curfans[i]];
for (int j = 0; j < fansize2; j++)
{
if (vis[cur[j]] == 0) {
tmp.push_back(cur[j]); vis[cur[j]] = 1;
}
}
}
curfans.clear();
curfans.resize(tmp.size());
copy(tmp.begin(), tmp.end(), curfans.begin());
bfs(l-1);
}
int main()
{
//freopen("input.txt", "r", stdin);
PAT *p = new PAT;
p->run();
return 0;
}