# 【題解】ZeroJudge e796: p3. 站牌廣告

【題目敘述】https://zerojudge.tw/ShowProblem?problemid=e796
【解題想法】

• (Line-22) 若有多個則輸出編號最大者。
```#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int ppl[1005];

int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int B, P, x, y;
cin >> B >> P;
for (int i=0; i<P; i++){
cin >> x >> y;
if (y < x) swap(x, y);
for (int j=x; j<=y; j++)
ppl[j]++;
}
int mn = 1e9, idx_mn = -1;
int mx = 0, idx_mx = -1;
for (int i=1; i<=B; i++){
if (ppl[i] >= mx) {
mx = ppl[i];
idx_mx = i;
}
if (ppl[i] < mn) {
mn = ppl[i];
idx_mn = i;
}
}
cout << idx_mn << ' ' << idx_mx << '\n';
return 0;
}
```

Python code (credit: Amy Chou)
⚠️注意：實際測資與範例格式不同，B、P分成兩行輸入。

```B = int(input())
P = int(input())
ppl = [0 for _ in range(B+1)]
for _ in range(P):
x, y = map(int, input().split())
if y < x:
x, y = y, x
for j in range(x, y+1):
ppl[j] += 1

mn = 1000000
mx = 0
idx_mx = -1
idx_mn = -1
for i in range(1, B+1):
if ppl[i] >= mx:
mx = ppl[i]
idx_mx = i
if ppl[i] < mn:
mn = ppl[i]
idx_mn = i

print(idx_mn, idx_mx)

```