博主分享了PAT编程竞赛中多个题目（如1001A+Bformat、1002A+BforPolynomials等）的解题思路和代码实现，涉及算法包括层次遍历、Dijkstra求最短路径、并查集、二分查找等。同时，博主表达了对于编程竞赛和未来职业规划的焦虑与期待，分享了收到名校录取通知的喜悦。
随便做做，不知道做啥了，看看没有讲解做成啥样
1001 A + B format
a加b然后按照千分位三个一组输出。
#include<iostream>
#include<vector>
using namespace std;
int main()
{
int a, b, c;
bool s = false;
cin >> a >> b;
c = a + b;
vector<int> v;
if(c < 0)
{
c = -c;
cout << '-';
}
int cnt = 0;
//从低位开始，凑够三位加一个逗号标记，然后正序输出
// 不断取最低位 while(c) { x = c % 10， c /= 10;}
while(c)
{
if(cnt == 3)
{
v.push_back(10);//逗号标记
cnt = 0;
}
v.push_back(c % 10);
cnt++;
c = c / 10;
}
for(int i = v.size() - 1; i >= 0; i--)
{
if(v[i] == 10)
cout << ',';
else cout << v[i];
}
if(v.size() == 0)
cout << 0;
return 0;
}
1002 A+B for Polynomials (25 分)
给了两个多项式的系数和下标，对应位相加即可。
最后要输出不是0的个数 和系数下标
#include<iostream>
using namespace std;
const int N = 1010;
int k, n, cnt;
double p[N], a;
int main()
{
scanf("%d", &k);
while(k--)
{
scanf("%d %lf", &n, &a);
if(p[n] == 0)
cnt++;
p[n] += a;
if(p[n] == 0)
cnt--;
}
scanf("%d", &k);
while(k--)
{
scanf("%d %lf", &n, &a);
if(p[n] == 0)
cnt++;//cnt实时统计不为0的个数，或者最后统一计算一遍也可
p[n] += a;
if(p[n] == 0)
cnt--;
}
printf("%d", cnt);
int c = 0;
for(int i = N - 1; i >= 0; i--)
{
if(p[i])
printf(" %d %.1f", i, p[i]);
}
return 0;
}
1003 Emergency (25 分) ※
dijsktra求单源最短路径，此题目的要点在于求最短路径的条数和最短路径上最大的队伍和
dist[t] + g[t][j] < dist[j] //t和j在一条最短路径上,路径不变，队伍数是前面的t再加上j点处的队伍数- dist[j] = dist[t] + g[t][j];
sum[j] = sum[t] + team[j];
num[j] = num[t];
- dist[j] = dist[t] + g[t][j];
dist[t] + g[t][j] == dist[j])//经过t到j和不经过t到j的最短路径相同, 这时最短路径就产生了分叉，路径条数相加,队伍数取最大- num[j] += num[t];
sum[j] = max(sum[j], sum[t] + team[j]);
- num[j] += num[t];
#include<iostream>
#include<cstring>
#include<set>
using namespace std;
const int N = 510;
int dist[N], g[N][N], team[N];
int sum[N], num[N]; //num[]表示从c2到各个点最短路径的条数, sum[]表示从c2到各个点最短路径上的队伍个数
bool st[N];
int n, m, c1, c2;
void dijsktra(int& shortest, int& cnt)
{
memset(dist, 0x3f, sizeof dist);
dist[c1] = 0;
sum[c1] = team[c1];
num[c1] = 1;//保证有一条
for(int i = 0; i < n; i++)
{
int t = -1;
for(int j = 0; j < n; j++)
{
if(!st[j] && (t == -1 || dist[j] < dist[t]))
t = j;
}
st[t] = true;
for(int j = 0; j < n; j++)
{
if(dist[t] + g[t][j] < dist[j]) //t和j在一条最短路径上,路径不变
{
dist[j] = dist[t] + g[t][j];
sum[j] = sum[t] + team[j];
num[j] = num[t];
}
else if(dist[t] + g[t][j] == dist[j])//经过t到j和不经过t到j的最短路径相同, 路径条数相加,队伍数取最大
{
num[j] += num[t];
sum[j] = max(sum[j], sum[t] + team[j]);
}
}
}
shortest = num[c2];
cnt = sum[c2];
}
int main()
{
int shortest, cnt;
memset(g, 0x3f, sizeof g);
scanf("%d%d%d%d", &n, &m, &c1, &c2);
for(int i = 0; i < n; i++)//是第i个城市拥有的城市数目而不是队伍编号
scanf("%d", &team[i]);
while(m--)
{
int a, b, w;
scanf("%d%d%d", &a, &b, &w);
g[a][b] = w;
g[b][a] = w;
}
dijsktra(shortest, cnt);
printf("%d %d", shortest, cnt);
return 0;
}
1004 Counting Leaves (30 分)
层次遍历法 求每一层叶节点的数目
#include<iostream>
#include<cstring>
#include<queue>
#include<unordered_map>
using namespace std;
const int N = 110;
int n, m, a, b, k;
int h[N], e[N], ne[N], idx;
void add(int a, int b)
{
e[idx] = b;
ne[idx] = h[a];
h[a] = idx;
idx++;
}
int main()
{
memset(h, -1, sizeof h);
unordered_map<int, int> hashset; //保存非叶节点，用unordered_set更好其实
scanf("%d %d", &n, &m);
while(m--)
{
scanf("%d %d", &a, &k);
hashset[a] = k;
while(k--)
{
scanf("%d", &b);
add(a, b);
}
}
queue<int> q;
q.push(1);
int flag = false;
while(q.size())
{
int si = q.size(), cnt = 0;
for(int i = 0; i < si; i++)
{
int t = q.front();
q.pop();
if(!hashset.count(t))
cnt++;//每层中统计一次个数
for(int i = h[t]; i != -1; i = ne[i])
{
int j = e[i];
q.push(j);
}
}
if(flag)
printf(" %d", cnt);
else
{
printf("%d", cnt);
flag = true;
}
}
return 0;
}
1005 Spell It Right (20 分)
算大数的各个数位之和
#include<iostream>
#include<vector>
using namespace std;
//各个位置数相加不会超过int
int main()
{
string a;
vector<int> A;
int sum = 0;
cin >> a;
for(int i = a.size() - 1; i >= 0; i--)
sum += a[i] - '0';
while(sum)
{
A.push_back(sum % 10);
sum /= 10;
}
//输出
string out[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int n = A.size();
for(int i = n - 1; i >= 0; i--)
{
if(i == n - 1)
cout << out[A[i]];
else
cout << " " << out[A[i]] ;
}
if(n == 0)
cout << "zero";
return 0;
}
1006 Sign In and Sign Out (25 分)
日期转换
#include<iostream>
#include<climits>
using namespace std;
int main()
{
int n, h1, m1, s1, h2, m2, s2;
string name;
cin >> n;
int mi = INT_MAX, ma = INT_MIN;
string unlock, lock;
while(n--)
{
cin >> name;
scanf("%d:%d:%d %d:%d:%d", &h1, &m1, &s1, &h2, &m2, &s2);
int t1 = h1 * 3600 + m1 * 60 + s1;
int t2 = h2 * 3600 + m2 * 60 + s2;
if(t1 < mi)
{
mi = t1;
unlock = name;
}
if(t2 > ma)
{
ma = t2;
lock = name;
}
}
cout << unlock << " " << lock;
return 0;
}
二次遍历-前缀和过了，无语总是看不清题意，我以为输出下标，结果是数
#include<iostream>
#include<climits>
using namespace std;
const int N = 1e4 + 10;
int a[N], sum[N];
int n, s, k;
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
sum[i] = sum[i - 1] + a[i];
//printf("%d ", sum[i]);
if(a[i] < 0)
k++;
}
int ma = INT_MIN, ml = 0, mr = 0;
for(int i = 1; i <= n; i++)
{
for(int j = i; j <= n; j++)
{
if(sum[j] - sum[i - 1] > ma)
{
ml = i, mr = j;
ma = sum[j] - sum[i - 1];
}
}
}
if(k == n)
printf("0 %d %d", a[1], a[n]);
else
printf("%d %d %d", ma, a[ml], a[mr]);
return 0;
}
一次遍历代码（柳神）※
#include<iostream>
#include<climits>
using namespace std;
const int N = 1e4 + 10;
int a[N];
int n, k;
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
if(a[i] < 0)
k++;
}
int ml = 1, mr = n, sum = INT_MIN, tmp = 0, tmpl = 1;
for(int i = 1; i <= n; i++)
{
tmp += a[i];
if(tmp < 0)//临时和小于0了，舍弃前面统计的这些，临时左下标tmpl从i的下一个开始
{
tmp = 0;
tmpl = i + 1;
}
else if(tmp > sum)//临时和大于sum了，更新此时和和位置
{
sum = tmp;
ml = tmpl;
mr = i;
}
}
if(k == n)
printf("0 %d %d", a[1], a[n]);
else
printf("%d %d %d", sum, a[ml], a[mr]);
return 0;
}
1008 Elevator (20 分)
20分的都太easy
#include<iostream>
using namespace std;
int main()
{
int n, f, t = 0, s = 0;
cin >> n;
while(n--)
{
cin >> f;
if(t < f)
s += (f - t) * 6;
else
s += (t - f) * 4;
s += 5;
t = f;
}
printf("%d", s);
return 0;
}
1009 Product of Polynomials (25 分)
乘法结果要开两倍
#include<iostream>
using namespace std;
const int N = 2010;
double a[N], c[N];
int n1, n2, x, cnt;
int main()
{
double k;
scanf("%d", &n1);
while(n1--)
{
scanf("%d %lf", &x, &k);
a[x] = k;
}
scanf("%d", &n2);
while(n2--)
{
scanf("%d %lf", &x, &k);
for(int i = 0; i < N; i++)
{
if(a[i])
c[i + x] += k * a[i];
}
}
for(int i = 0; i < N; i++)
{
if(c[i])
cnt++;
}
printf("%d", cnt);
for(int i = N - 1; i >= 0; i--)
{
if(c[i])
printf(" %d %.1f", i, c[i]);
}
return 0;
}
1010 Radix (25 分) ※
有点难，主要是情况复杂需要考虑清楚
y总二分模板yyds！
#include<iostream>
#include<ctype.h>
#include<algorithm>
using namespace std;
long long convert(string s, long long r)
{
long long ans = 0;
for(int i = 0; i < s.size(); i++)
{
int t = isdigit(s[i])? s[i] - '0': s[i] - 'a' + 10;
ans = ans * r + t;
}
return ans;
}
int main()
{
string n1, n2, x;
int tag;
long long radix, N1, N2;
cin >> n1 >> n2 >> tag >> radix;
if(tag == 2) //n1为已知进制数字,n2为未知进制数字
x = n1, n1 = n2, n2 = x;
N1 = convert(n1, radix);
char ma_char = '0'; //求出n2中最大字符
for(int i = 0; i < n2.size(); i++)
if(n2[i] > ma_char)
ma_char = n2[i];
int ma_num = isdigit(ma_char)? ma_char - '0': ma_char - 'a' + 10;
//96549157373046880 10 1 10，输出应该是96549157373046880
//n1可能很大,导致上届很大,进制很大,转换就可能越界long long变成负数
long long l = ma_num + 1;//左边界最小为x中最大字符+ 1 ab:最小为c,12进制
long long r = max(l, N1); //x不能超过N1,最小两位数10的最大进制只能到N1进制,转换成十进制这就已经是N1了,再大的进制都不可能了
//一位数的话,可能n1比n2的最小进制还要小,比如n1=1, n2=1, radix = 2,最小进制是2,和n1取max
while(l < r)
{
long long mid = l + r + 1 >> 1;
N2 = convert(n2, mid);
if(N2 < 0 || N2 > N1)
r = mid - 1;//false
else
l = mid;//right
}
if(N1 == convert(n2, l))
printf("%d", l);
else
printf("Impossible");
return 0;
}
1011 World Cup Betting (20 分)
题目看不懂，没意思的一道题
#include<iostream>
using namespace std;
int main()
{
char g[3] = {'W', 'T', 'L'};
double x, ans = 1;
for(int i = 0; i < 3; i++)
{
double ma = 0;
int idx = 0;
for(int j = 0; j < 3; j++)
{
scanf("%lf", &x);
if(x > ma)
{
ma = x;
idx = j;
}
}
printf("%c ", g[idx]);
ans = ans * ma;
}
printf("%.2f", (ans * 0.65 - 1) * 2);
return 0;
}
下面先只做25和30分的非模拟题
1013 Battle Over Cities (25 分)
并查集算连通块数量做的，垃圾为啥不给M范围，测试点4边巨大。看来是个稠密图，用邻接矩阵快一些。
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1010;
int n, m, k, a, b, x;
int g[N][N], father[N];
int find(int x)
{
if(father[x] != x)
father[x] = find(father[x]);
return father[x];
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
memset(g, 0x3f, sizeof g);
while(m--)
{
scanf("%d%d", &a, &b);
g[a][b] = 1;
g[b][a] = 1;
}
while(k--)
{
scanf("%d", &x);
int cnt = n - 1;
for(int i = 1; i <= n; i++)
father[i] = i;
for(int i = 1; i <= n; i++)
{
if(i == x) continue;
for(int j = 1; j <= n; j++)
{
a = i, b = e[j];
//cout << a << b << endl;
if(b == x || g[a][b] == 0x3f3f3f3f) continue;
int fa = find(a), fb = find(b);
if(fa != fb)
{
father[fa] = fb;
cnt--;
}
}
}
if(k == 0)
printf("%d", cnt - 1);
else
printf("%d\n", cnt - 1);
}
return 0;
}
柳神DFS思路：n次迭代，看能dfs几次，就说明有几个不连通，nice！
#include<iostream>
#include<cstring>
using namespace std;
const int N = 1010;
int n, m, k, a, b, x;
int g[N][N];
bool st[N];
void dfs(int x)
{
st[x] = true;
for(int i = 1; i <= n; i++)
if(!st[i] && g[x][i] == 1)
dfs(i);
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
memset(g, 0x3f, sizeof g);
while(m--)
{
scanf("%d%d", &a, &b);
g[a][b] = 1;
g[b][a] = 1;
}
while(k--)
{
scanf("%d", &x);
int cnt = 0;
memset(st, false, sizeof st);
st[x] = true;
for(int i = 1; i <= n; i++)
if(!st[i])
{
dfs(i);cnt++;
}
if(k == 0)
printf("%d", cnt - 1);
else
printf("%d\n", cnt - 1);
}
return 0;
}
1015 Reversible Primes (20 分)
想着复习一下素数筛法
进制转换 + 素数判断
#include<iostream>
#include<vector>
using namespace std;
const int N = 1e7;
bool st[N];
//埃式筛法
void get_prime()
{
st[1] = true, st[0] = true;//测试点2
for(int i = 2; i < N; i++)
{
if(!st[i])
{
for(int j = i + i; j < N; j = j + i)
st[j] = true;
}
}
}
int main()
{
int n, r;
get_prime();
while(1)
{
scanf("%d", &n);
if(n < 0)
break;
scanf("%d", &r);
if(st[n])
{
printf("No\n");
continue;
}
vector<int> inverse;
//从十进制转换成r进制
while(n)
{
inverse.push_back(n % r);
n = n / r;
}
//从r进制转换成十进制
int in = 0;
for(int i = 0; i < inverse.size(); i++)
{
in = in * r + inverse[i];
}
//cout << in << endl;
if(!st[in])
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
线性筛法
int prime[N], cnt;
bool st[N];
void get_prime()
{
st[1] = true;
for(int i = 2; i < N; i++)
{
if(!st[i])
prime[cnt++] = i;
for(int j = 0; prime[j] < N / i; j++)
{
st[prime[j] * i] = true;
if(i % prime[j] == 0)
break;
}
}
}
1018 Public Bike Management (30 分)
累了这道题
首先dijkstra找到最短路径的前驱节点
- if(dist[t] + g[t][j] < dist[j])
- pre[j].clear();
pre[j].push_back(t);
- pre[j].clear();
- else if(dist[t] + g[t][j] == dist[j])
- pre[j].push_back(t);
dfs统计mi_need和mi_back
真的很恶心，不能用总的-完美的，因为不能用后面的补前面，所以还是得一个一个来！
#include<iostream>
#include<cstring>
#include<climits>
#include<vector>
using namespace std;
const int N = 510;
int bike[N], dist[N], b[N];
int g[N][N];
bool st[N];
vector<vector<int>> pre(N);
vector<int> tmppath, path;
int cm, n, m, sp, u, v, w;
int sum = 0, cnt = 0, mi_need = INT_MAX, mi_back = INT_MAX;
//记录前驱节点
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
dist[0] = 0;
for(int i = 0; i <= n; i++)
{
int t = -1;
for(int j = 0; j <= n; j++)
{
if(!st[j] && (t == -1 || dist[t] > dist[j]))
t = j;
}
st[t] = true;
for(int j = 0; j <= n; j++)
{
if(dist[t] + g[t][j] < dist[j])
{
dist[j] = min(dist[j], dist[t] + g[t][j]);
pre[j].clear();
pre[j].push_back(t);
}
else if(dist[t] + g[t][j] == dist[j])
{
dist[j] = min(dist[j], dist[t] + g[t][j]);
pre[j].push_back(t);
}
}
}
return dist[sp];
}
void dfs(int u)
{
tmppath.push_back(u);
if(u == 0)
{
int prefect = cm / 2 , back = 0, need = 0;
//可以用前面缺的车补后面,但不能用后面补前面
for(int i = tmppath.size() - 2; i >= 0; i--)
{
v = tmppath[i];
if(bike[v] < prefect)
{
if(back > prefect - bike[v])
back -= (prefect - bike[v]);
else
{
need += ((prefect - bike[v]) - back);
back = 0;
}
}
else
back += (bike[v] - prefect);
}
if(need < mi_need)
{
mi_need = need;
path = tmppath;
mi_back = back;
}
else if(need == mi_need && back < mi_back)
{
mi_back = back;
path = tmppath;
}
tmppath.pop_back();//pop 0
return ;
}
for(int i = 0; i < pre[u].size(); i++)
dfs(pre[u][i]);
tmppath.pop_back();
}
int main()
{
cin >> cm >> n >> sp >> m;
for(int i = 1; i <= n; i++)
scanf("%d", &bike[i]);
memset(g, 0x3f, sizeof g);
while(m--)
{
scanf("%d %d %d", &u, &v, &w);
g[u][v] = g[v][u] = w;
}
int x = dijkstra();
//tmppath.push_back(sp);
dfs(sp);
printf("%d 0", mi_need);
for(int i = path.size() - 2; i >= 0; i--)
printf("->%d", path[i]);
printf(" %d", mi_back);
return 0;
}
靠 ! 没保存!
1020 Tree Traversals (25 分)
中序+ 后序构建二叉树,然后层次遍历
easy
#include<iostream>
#include<queue>
using namespace std;
const int N = 35;
int in[N], out[N], ans[N];
int n, idx;
struct Node
{
int val;
Node* left;
Node* right;
Node(int v)
{
val = v;
left = right = nullptr;
}
};
Node* construct(int li, int ri, int lo, int ro)
{
if(li > ri || lo > ro)
return nullptr;
Node* root = new Node(out[ro]);
int inid = 0;//中序中rootid
for(int i = li; i <= ri; i++)
{
if(out[ro] == in[i])
{
inid = i;
break;
}
}
int leftsize = inid - li, rightsize = ri - inid;
root->left = construct(li, inid - 1, lo, lo + leftsize - 1);
root->right = construct(inid + 1, ri, lo + leftsize, ro - 1);
return root;
}
void bfs(Node * root)
{
queue<Node*> q;
q.push(root);
while(q.size())
{
Node * t = q.front();
q.pop();
ans[idx++] = t->val;
if(t->left != nullptr)
q.push(t->left);
if(t->right != nullptr)
q.push(t->right);
}
}
int main()
{
cin >> n;
for(int i = 1; i <= n; i++)
scanf("%d", &out[i]);
for(int i = 1; i <= n; i++)
scanf("%d", &in[i]);
Node* root = construct(1, n, 1, n);
bfs(root);
printf("%d", ans[0]);
for(int i = 1; i < n; i++)
printf(" %d", ans[i]);
return 0;
}
1021 Deepest Root (25 分)
找到根最深的root
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int N = 100010;
int h[N], e[N], ne[N], idx;
int n, u, v, cnt = 0;
bool st[N];
int height[N];
void add(int u, int v)
{
e[idx] = v; ne[idx] = h[u]; h[u] = idx; idx++;
}
void dfs(int u)
{
st[u] = true;
for(int i = h[u]; i != -1; i = ne[i])
{
int j = e[i];
if(!st[j])
dfs(j);
}
}
int bfs(int u)
{
memset(st, false, sizeof st);
queue<int> q;
q.push(u);
st[u] = true;
int ans = 1;
while(q.size())
{
int qn = q.size();
for(int i = 0; i < qn; i++)
{
int t = q.front();
q.pop();
for(int j = h[t]; j != -1; j = ne[j])
{
v = e[j];
if(!st[v])
{
q.push(v);
st[v] = true;
}
}
}
ans++;
}
return ans;
}
int main()
{
scanf("%d", &n);
memset(h, -1, sizeof h);
for(int i = 1; i < n; i++)
{
scanf("%d %d", &u, &v);
add(u, v);
add(v, u);
}
//DFS找连通块数量
for(int i = 1; i <= n; i++)
{
if(!st[i])
{
dfs(i);
cnt++;
}
}
if(cnt > 1)
{
printf("Error: %d components", cnt);
return 0;
}
//BFS找高度最大的
int ma = 0;
for(int i = 1; i <= n; i++)
{
height[i] = bfs(i);
ma = max(ma, height[i]);
}
for(int i = 1; i <= n; i++)
if(height[i] == ma)
printf("%d\n", i);
return 0;
}
1022 Digital Library (30 分)
最后一个测试点没过
#include<iostream>
#include<sstream>
using namespace std;
#include<map>
#include<set>
int main()
{
int n, m;
string line, id, year, title, author, keyword, publisher;
multimap<string, string> myear, mtitle, mauthor, mkeyword, mpublisher;
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
cin >> id;
getchar();
getline(cin, title);
getline(cin, author);
getline(cin, keyword);
getline(cin, publisher);
cin >> year;
getchar();
mtitle.insert(make_pair(title, id));
myear.insert(make_pair(year, id));
mauthor.insert(make_pair(author, id));
istringstream skeyword(keyword);
string word;
while(skeyword >> word)
mkeyword.insert(make_pair(word, id));
mpublisher.insert(make_pair(publisher, id));
}
scanf("%d", &m);
getchar();
while(m--)
{
getline(cin, line);
cout << line << endl;
string str = line.substr(3, line.size() - 3);
switch (line[0])
{
case '1':
{
auto nums = mtitle.count(str);
auto iter = mtitle.find(str);
if(nums == 0) cout << "Not Found\n";
while(nums--)
{
cout << iter->second << endl;
iter++;
}
break;
}
case '2':
{
auto nums = mauthor.count(str);
auto iter = mauthor.find(str);
if(nums == 0) cout << "Not Found\n";
while(nums--)
{
cout << iter->second << endl;
iter++;
}
break;
}
case '3':
{
auto nums = mkeyword.count(str);
auto iter = mkeyword.find(str);
if(nums == 0) cout << "Not Found\n";
set<string> setkey;
while(nums--)
{
setkey.insert(iter->second);
iter++;
}
for(auto tmp: setkey)
cout << tmp << endl;
break;
}
case '4':
{
auto nums = mpublisher.count(str);
auto iter = mpublisher.find(str);
if(nums == 0) cout << "Not Found\n";
while(nums--)
{
cout << iter->second << endl;
iter++;
}
break;
}
case '5':
{
auto nums = myear.count(str);
auto iter = myear.find(str);
if(nums == 0) cout << "Not Found\n";
while(nums--)
{
cout << iter->second << endl;
iter++;
}
break;
}
default:
break;
}
}
return 0;
}
1024 Palindromic Number (25 分)
回文判断 + 大整数加法
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 50;
bool check(vector<int>& a)
{
int an = a.size();
int i = 0, j = an - 1;
while(i < j) //不能等于 可能i > j
{
if(a[i] != a[j])
return false;
i++, j--;
}
return true;
}
vector<int> add(vector<int>& a, vector<int> & b)
{
vector<int> c;
int an = a.size(), bn = b.size();
int t = 0;
for(int i = 0; i < an || i < bn; i++)
{
if(i < an)
t += a[i];
if(i < bn)
t += b[i];
c.push_back(t % 10);
t = t / 10;
}
if(t)
c.push_back(t);
return c;
}
int main()
{
string a;
int k, cnt = 0;
vector<int> A;
cin >> a >> k;
for(int i = a.size() - 1; i >= 0; i--)
A.push_back(a[i] - '0');
while(!check(A) && cnt < k)
{
vector<int> B;
for(int i = A.size() - 1; i >= 0; i--)
B.push_back(A[i]);
A = add(A, B);
cnt++;
}
for(int i = A.size() - 1; i >= 0; i--)
printf("%d", A[i]);
cout << endl << cnt;
return 0;
}
1028 List Sorting (25 分)
结构体+ sort排序 easy
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
struct stu
{
string id, name;
int grade;
}Stu[N];
bool cmp1(const stu& a, const stu& b)
{
return a.id < b.id;
}
bool cmp2(const stu& a, const stu& b)
{
if(a.name != b.name)
return a.name < b.name;
return a.id < b.id;
}
bool cmp3(const stu& a, const stu& b)
{
if(a.grade != b.grade)
return a.grade < b.grade;
return a.id < b.id;
}
int main()
{
int n, c;
cin >> n >> c;
for(int i = 0; i < n; i++)
cin >> Stu[i].id >> Stu[i].name >> Stu[i].grade;
switch (c)
{
case 1:
sort(Stu, Stu + n, cmp1);
break;
case 2:
sort(Stu, Stu + n, cmp2);
break;
case 3:
sort(Stu, Stu + n, cmp3);
break;
default:
break;
}
for(int i = 0; i < n; i++)
cout << Stu[i].id << " " << Stu[i].name << " " << Stu[i].grade << endl;
return 0;
}
28号推免结束后再见 ! 希望希望! 好运好运! 再给我最后一次好运吧 ! !
最近总是焦虑的不行，因为我非常清楚自己的职业规划，甚至在每一步都想好了大概要做什么。但是当你达不到或者有机会但是发现太难了的时候，你就会陷入深深的焦虑、难过与不自信中。都到这个时候了，保研都进行两个月了。该做的都做得差不多了，剩下的就是别紧张，放轻松，能行就行不行就算了吧，可能自己的水平还不够，下一个阶段在努力吧！
28号见！
————————————————————————————————————————
29号更新！幸运来得太突然，接连收到浙大、复旦、北大软微的复试通知，由于冲突，选择二度放弃复旦面试。最终获得了浙大、北大软微的offer，去北京啦！PAT也没太用上！但是我相信一切都是缘分！感谢自己！感谢大家！PAT也就不会再更新啦！祝大家心想事成嘿嘿！！！
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