程式碼
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int main()
{
string s;
int n;
cin >> s >> n;
int i = 0;
while (i != s.size())
{
if (s[i] - '0' == n)
{
s.erase(i, 1);
}
else
i++;
}
bool flag = true;
for (int i = 0; i < s.size() / 2; i++)
{
if (s[i] != s[s.size() - i - 1])
{
flag = false;
break;
}
}
if (flag) cout << "Yes";
else cout << "No";
return 0;
}
解題心得
題目中的利潤跟我認知的不一樣，整題花在理解題意上面的時間最久。
程式碼
#include<iostream>
using namespace std;
int main()
{
int n, d, a[101] = { 0 }, sum = 0, current = -1;
bool flag = false;
cin >> n >> d;
for (int i = 0; i < n; i++)
{
cin >> a[i];
if (i == 0)
{
current = a[i];
flag = true;
}
else if (flag && a[i] >= current + d)
{
sum += a[i] - current;
current = a[i];
flag = false;
}
else if (!flag && a[i] <= current - d) // 當下沒有持有股票
{
current = a[i];
flag = true;
}
}
cout << sum;
return 0;
}
程式碼
#include<iostream>
using namespace std;
int main()
{
int n, file[100] = { 0 }, index = 0;
string s;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> s;
index = (s[3] - '0') * 10 + (s[4] - '0');
file[index] ++ ;
}
for (int i = 0; i < 100; i++)
{
if (file[i])
{
cout << i << " " << file[i] << endl;
}
}
return 0;
}
程式碼
#include<iostream>
using namespace std;
int main()
{
int index[10] = { 0 }, maxTime = 0, n;
for (int i = 0; i < 3; i++)
{
cin >> n;
index[n]++;
}
for (int i = 0; i < 10; i++)
{
if (index[i] > maxTime)
maxTime = index[i];
}
cout << maxTime << " ";
for (int i = 9; i >= 0; i--)
{
if (index[i])
{
cout << i;
maxTime--;
if (3 - maxTime != 0) cout << " ";
}
}
return 0;
}
解題心得
往左看一次，往右也看一次。
程式碼
#include<iostream>
using namespace std;
int main()
{
int n, m, arr[1001], sum = 0, height;
cin >> n >> m;
for (int i = 0; i < n; i++)
cin >> arr[i];
m -= 1;
height = arr[m];
for (int i = m - 1; i >= 0; i--)
{
if (arr[i] > height)
{
height = arr[i];
sum++;
}
}
height = arr[m];
for (int i = m + 1; i < n; i++)
{
if (arr[i] > height)
{
height = arr[i];
sum++;
}
}
cout << sum;
return 0;
}
解題心得
想了一下才想到如何判斷區間QQ
程式碼
#include<iostream>
using namespace std;
int main()
{
int k, w, s, e, sum = 0;
cin >> k >> w >> s >> e;
if (k < 2)
sum += 20;
else
sum += 20 + (k - 2) * 5;
sum += (w / 2) * 5;
bool time[24] = { false };
for (int i = s; i <= e; i++)
time[i] = true;
if (time[18] && time[19]) sum += 185;
if (time[19] && time[20]) sum += 195;
if (time[20] && time[21]) sum += 205;
if (time[21] && time[22]) sum += 215;
if (time[22] && time[23]) sum += 225;
cout << sum;
return 0;
}
程式碼
#include<iostream>
using namespace std;
int main()
{
int x, y, n, fishX[501] = { 0 }, fishY[501] = { 0 }, ansX = 0, ansY = 0;
cin >> x >> y >> n;
for (int i = 0; i < n; i++)
{
cin >> fishX[i] >> fishY[i];
int now = abs(x - fishX[i]) * abs(x - fishX[i]) + abs(y - fishY[i]) * abs(y - fishY[i]);
int best = abs(x - fishX[ansX]) * abs(x - fishX[ansX]) + abs(y - fishY[ansY]) * abs(y - fishY[ansY]);
if (now < best)
{
ansX = i;
ansY = i;
}
}
cout << fishX[ansX] << " " << fishY[ansY];
return 0;
}
解題心得
原本left<=right沒有加等號，加上後就過了。
程式碼
#include<iostream>
#include<vector>
using namespace std;
int main()
{
long long int n, q, x, t;
vector<long long int> v;
cin >> n >> q;
for (int i = 0; i < n; i++)
{
cin >> x;
v.push_back(x);
}
while (q--)
{
cin >> t;
int left = 0, right = v.size() - 1, mid = (left + right) / 2;
bool flag = false;
while (left <= right)
{
if (v[mid] == t)
{
flag = true;
break;
}
else if (v[mid] < t)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
mid = (left + right) / 2;
}
if (flag) cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
}
解題心得
跟這題有點像。
程式碼
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
string s1, s2;
int base = 0, col1 = 0, col2 = 0;
cin >> s1 >> s2;
for (int i = s1.size() - 1; i >= 0; i--)
{
col1 += (s1[i] - 'A' + 1) * pow(26, base);
base++;
}
base = 0;
for (int i = s2.size() - 1; i >= 0; i--)
{
col2 += (s2[i] - 'A' + 1) * pow(26, base);
base++;
}
cout << abs(col1 - col2) + 1;
return 0;
}
解題心得
AAA3 = AAA(col) * 3(row)
AAA = A * 26^2 + A * 26^1 + A
程式碼
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
string s;
int col = 0, row = 0, base = 1;
cin >> s;
while (isdigit(s.back()))
{
row += base * (s.back() - '0');
s.pop_back();
base *= 10;
}
base = 0;
for (int i = s.size() - 1; i >= 0; i--)
{
col += (s[i] - 'A' + 1) * pow(26, base);
base++;
}
cout << col * row;
return 0;
}
程式碼
#include<iostream>
using namespace std;
int main()
{
int h, w, r1, c1, r2, c2;
cin >> h >> w >> r1 >> c1 >> r2 >> c2;
r1 -= 1; c1 -= 1; r2 -= 1; c2 -= 1;
char puzzle[20][50];
for (int i = 0; i < h; i++)
for (int j = 0; j < w; j++)
cin >> puzzle[i][j];
if (r1 == r2)
{
for (int i = c1; i <= c2; i++)
cout << puzzle[r1][i];
}
else if (c1 == c2)
{
for (int i = r1; i <= r2; i++)
cout << puzzle[i][c1];
}
else
{
for (int i = 0; i <= abs(r1 - r2); i++)
cout << puzzle[r1 + i][c1 + i];
}
return 0;
}
程式碼
#include<iostream>
using namespace std;
int main()
{
string ans, s;
int m;
cin >> ans >> m;
while (m--)
{
cin >> s;
int a = 0, b = 0;
for (int i = 0; i < s.size(); i++)
{
for (int j = 0; j < s.size(); j++)
{
if (ans[i] == s[j])
{
if (i == j) a++;
else b++;
}
}
}
cout << a << "A" << b << "B" << endl;
}
return 0;
}
解題心得
用stack，把字串倒著看。
程式碼
#include<iostream>
#include<stack>
#include<vector>
#include <string>
using namespace std;
int main()
{
int x, y, z;
stack<int> st;
vector<string> v;
string s;
while (cin >> s)
v.push_back(s);
for (int i = v.size() - 1; i >= 0; i--)
{
if (v[i] == "f")
{
x = st.top(); st.pop();
x = 2 * x - 3;
st.push(x);
}
else if (v[i] == "g")
{
x = st.top(); st.pop();
y = st.top(); st.pop();
x = 2 * x + y - 7;
st.push(x);
}
else if (v[i] == "h")
{
x = st.top(); st.pop();
y = st.top(); st.pop();
z = st.top(); st.pop();
x = 3 * x - 2 * y + z;
st.push(x);
}
else
{
st.push(stoi(v[i]));
}
}
cout << st.top();
return 0;
}
程式碼
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int n, t, brownie[1001] = { 0 }, info[1001] = { 0 }, sum = 0;
bool isVisited[1001] = { false };
cin >> n >> t;
for (int i = 0; i < n; i++)
cin >> info[i];
for (int i = 0; i < n; i++)
cin >> brownie[i];
while (!isVisited[t])
{
sum += brownie[t];
isVisited[t] = true;
t = info[t];
}
cout << sum;
return 0;
}
解題心得
照著題目敘述即可。
程式碼
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int n, k, t, arr[1001] = { 0 };
bool flag = false;
cin >> n >> k >> t;
for (int r = 0; r < n; r++)
{
for (int i = 0; i < k; i++)
cin >> arr[i];
sort(arr, arr + k);
int sum = 0;
for (int i = 1; i < k - 1; i++)
sum += arr[i];
if ((double)sum / (k - 2) >= t)
{
cout << r << endl;
flag = true;
}
}
if (!flag) cout << "A is for apple." << endl;
return 0;
}
解題心得
照著題目說明做就好了。
程式碼
#include <iostream>
using namespace std;
int main()
{
int x, r, v, n, p, s;
cin >> x >> r >> v >> n;
while (n--)
{
cin >> p >> s;
if (x - r <= p && p <= x + r && v >= s)
x = p;
else if (x - r <= p && p <= x + r && v < s)
{
if (p < x) x += 15;
else x -= 15;
}
}
cout << x;
return 0;
}
解題心得
事先算好，直接輸出。
程式碼
#include <iostream>
using namespace std;
int main()
{
long long int x[201] = { 0, 1 };
for (int i = 2; i <= 200; i++)
{
x[i] = x[i - 1] + i * i - i + 1;
}
int n;
cin >> n;
cout << x[n];
return 0;
}
解題心得
就很基本的stack(?
程式碼
#include <iostream>
#include <stack>
using namespace std;
int main()
{
int n, k, x;
stack<int> st;
cin >> n;
while (n--)
{
cin >> k;
if (k == 1)
{
cin >> x;
st.push(x);
}
else if (k == 2)
{
if (st.empty()) cout << -1 << endl;
else cout << st.top() << endl;
}
else if (k == 3)
{
if (!st.empty())
st.pop();
}
}
return 0;
}