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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#131056 | #2960. Social Distancing | l1924365846 | WA | 1ms | 4660kb | C++14 | 2.2kb | 2023-07-26 09:06:37 | 2023-07-26 09:06:40 |
Judging History
answer
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#include <unordered_map>
#include <unordered_set>
#define ll long long
#define int ll
#define ms(a,b) memset((a),(b),sizeof(a))
#define mc(a,b) memcpy((a),(b),sizeof(a))
#define PI acos(-1)
#define ssin(x) sin(x*PI/180.00)
#define ccos(x) cos(x*PI/180.00)
#define ttan(X) tan(x*PI/180.00)
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, n, a) for (int i = n - 1; i >= a; i--)
#define IOS ios::sync_with_stdio(false)
#define si(x) scanf("%d",&x)
#define sii(x, y) scanf("%d %d",&x, &y)
#define siii(x, y, z) scanf("%d %d %d",&x, &y, &z)
#define siiii(x, y, z, w) scanf("%d %d %d %d",&x, &y, &z, &w)
#define sl(x) scanf("%lld",&x)
#define sll(x, y) scanf("%lld %lld",&x, &y)
#define slll(x, y, z) scanf("%lld %lld %lld",&x, &y, &z)
#define sllll(x, y, z, w) scanf("%lld %lld %lld %lld",&x, &y, &z, &w)
#define pr(x) printf("%d",x)
#define pl(x) printf("%lld",x)
#define sd(x) scanf("%lf",&x)
#define sdd(x, y) scanf("%lf %lf",&x, &y)
#define sddd(x, y, z) scanf("%lf %lf %lf",&x, &y, &z)
#define sdddd(x, y, z, w) scanf("%lf %lf %lf %lf",&x, &y, &z, &w)
#define x first
#define y second
#define pb push_back
#define mk make_mair
using namespace std;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int mod = 998244353;
const int P = 131;
const int N = 1e5 + 10, M = 2e6 + 10;
const int inf = 0x3f3f3f3f;
const ll lnf = 9e18;
const double eps = 1e-6;
void solve() {
int s, n, res = 0;
int a[N];
sll (s, n);
for (int i = 0; i < n; i ++) {
sl (a[i]);
}
sort (a, a + n);
res += (a[0] - 1) / 2;
int l = a[0];
for (int i = 1; i < n; i ++) {
res += (a[i] - l - 2) / 2;
l = a[i];
}
if (a[0] % 2) {
res += (s - l - 1) / 2;
}
else {
res += (s - l) / 2;
}
if (a[0] % 2 && a[n - 1] % 2)
res--;
printf ("%lld\n", res);
}
signed main() {
int T (1);
// IOS;
// cin.tie (0);
// cout.tie (0);
// cin >> T;
// sl (T);
while ( T-- ) {
solve();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 4652kb
input:
9 2 2 6
output:
2
result:
ok single line: '2'
Test #2:
score: -100
Wrong Answer
time: 1ms
memory: 4660kb
input:
10 3 1 4 7
output:
0
result:
wrong answer 1st lines differ - expected: '1', found: '0'