Codeforces Round 918 (Div. 4)（补题AK）其实这次比赛没打算打的，看看题算了，但我的牛魔舍友在那bark，bark的叫，所以我就写了一道D题
A - Odd One Out
思路：简单题，if-else练习题
AC Code
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
map<int, int> w;
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int a, b, c;
w.clear();
scanf("%d%d%d", &a, &b, &c);
w[a]++;
w[b]++;
w[c]++;
if (w[a] == 1)
{
printf("%d\n", a);
}
else if (a == c)
{
printf("%d\n", b);
}
else
{
printf("%d\n", c);
}
}
return 0;
}
B - Not Quite Latin Square
思路：读入一个3*3的矩阵，每个都应出现3次，把少的那个输出即可
AC Code
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
char a[10][10];
map<char, int> cnt;
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
cnt.clear();
f(i, 0, 2)
{
scanf("%s", a[i]);
}
char tmp = 'A';
f(i, 0, 2)
{
f(j, 0, 2)
{
cnt[a[i][j]]++;
}
}
f(i, 0, 2)
{
if (cnt['A' + i] < 3)
{
printf("%c\n", 'A' + i);
break;
}
}
}
return 0;
}
C - Can I Square?
思路：简单的模拟，注意sqrt的精度即可
AC Code
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
ll a;
map<ll, int> cnt;
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
ll sum = 0;
ll tmp;
f(i, 1, n)
{
scanf("%lld", &tmp);
sum += tmp;
}
ll w = sqrt(sum);
int flag = 0;
for (ll i = w - 1; i <= w + 1; i++)
{
if (i < 0)
{
continue;
}
if (i * i == sum)
{
printf("YES\n");
flag = 1;
break;
}
}
if (flag == 0)
{
printf("NO\n");
}
}
return 0;
}
D - Unnatural Language Processing
思路：简单观察一下，得出规律即可
AC Code
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
char a[maxn];
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
getchar();
f(i, 1, n)
{
scanf("%c", &a[i]);
}
f(i, 1, n)
{
if (a[i] == 'b' || a[i] == 'c' || a[i] == 'd')
{
printf("%c", a[i]);
i++;
if (i == n)
{
printf("%c\n", a[i]);
break;
}
printf("%c", a[i]);
i++;
if (i == n)
{
printf("%c\n", a[i]);
break;
}
if (a[i + 1] == 'b' || a[i + 1] == 'c' || a[i + 1] == 'd')
{
printf("%c", a[i]);
printf(".");
}
else
{
printf(".");
i--;
}
}
}
}
return 0;
}
E - Romantic Glasses
思路：其实就是所有奇数乘一个负一就行了
AC Code
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
int a[maxn] = {0};
ll pre[maxn] = {0};
map<ll, int> w;
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
int flag = 0;
w.clear();
f(i, 1, n)
{
scanf("%d", &a[i]);
if (i % 2 == 1)
{
a[i] *= (-1);
}
pre[i] = pre[i - 1] + a[i];
}
f(i, 1, n)
{
w[pre[i]]++;
if (w[pre[i]] >= 2 || pre[i] == 0)
{
flag = 1;
break;
}
}
if (flag == 1)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
return 0;
}
F - Greetings
思路：先将左端点排升序，保证每一人都只会被后面序列的人撞到，其实就是求后面b[j]>b[i]的个数，也就是求逆序对的个数，把板子代入即可（排序问题的衍生问题）
AC Code
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
pair<int, int> a[maxn];
int b[maxn];
int tmp[maxn];
ll merge_sort(int q[], int l, int r)
{
if (l >= r)
{
return 0;
}
int mid = (l + r) >> 1;
ll res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
int i = l, j = mid + 1, k = 0;
while (i <= mid && j <= r)
{
if (q[i] <= q[j])
tmp[k++] = q[i++];
else
{
res += mid - i + 1;
tmp[k++] = q[j++];
}
}
while (i <= mid)
{
tmp[k++] = q[i++];
}
while (j <= r)
{
tmp[k++] = q[j++];
}
for (int i = l, j = 0; i <= r; i++, j++)
{
q[i] = tmp[j];
}
return res;
}
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
f(i, 1, n)
{
scanf("%d%d", &a[i].first, &a[i].second);
}
sort(a + 1, a + n + 1);
f(i, 1, n)
{
b[i] = a[i].second;
}
ll ans = merge_sort(b, 1, n);
printf("%lld\n", ans);
}
return 0;
}
G - Bicycles
思路：这应该就是最短路的板子题，我就学了一下迪杰斯特拉（Dijkstra)算法，也能做出来，然后不得不说的是这个填表的过程，总给我一种dp的感觉（哦，还有一个事情，千万不要用大顶堆当消愁）
AC Code
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define int long long
#define f(i, s, e) for (int i = s; i <= e; i++)
#define ff(i, s, e) for (int i = s; i >= e; i--)
#define maxn (ll)(2e5 + 5000)
using namespace std;
const ll linf = 5e18;
const ll N = 1100;
struct node
{
int num;
int dis;
int s;
bool operator>(const node &other) const
{
return dis > other.dis;
}
};
struct node tmp, fuck;
int n, m;
int dis[N][N];
int vis[N][N];
int cost[N];
vector<node> temp[N];
void dijkstra(int s)
{
f(i, 1, n)
{
f(j, 0, 1000)
{
dis[i][j] = linf;
}
}
priority_queue<node, vector<node>, greater<node>> q;
q.push({s, 0, cost[1]});
dis[s][cost[1]] = 0;
while (!q.empty())
{
tmp = q.top();
int x = tmp.num;
int y = tmp.s;
q.pop();
if (vis[x][y] == 1)
{
continue;
}
vis[x][y] = 1;
f(i, 0, (int)temp[x].size() - 1)
{
fuck = temp[x][i];
int len = temp[x][i].dis;
int next = temp[x][i].num;
int nexts = min(y, cost[next]);
if (dis[next][nexts] > dis[x][y] + len * y)
{
dis[next][nexts] = dis[x][y] + len * y;
q.push({next, dis[next][nexts], nexts});
}
}
}
}
signed main(void)
{
int t;
scanf("%lld", &t);
while (t--)
{
scanf("%lld%lld", &n, &m);
f(i, 1, n)
{
temp[i].clear();
f(j, 1, 1001)
{
vis[i][j] = 0;
}
}
f(i, 1, m)
{
int u, v, dist;
scanf("%lld%lld%lld", &u, &v, &dist);
temp[u].push_back({v, dist, 0});
temp[v].push_back({u, dist, 0});
}
f(i, 1, n)
{
scanf("%lld", &cost[i]);
}
dijkstra(1);
int ans = linf;
f(i, 0, 1000)
{
ans = min(ans, dis[n][i]);
}
printf("%lld\n", ans);
}
return 0;
}
总结：这个比赛毕竟是cf的最简单的，难得有一次能补完题，这就是今年最后一篇博客啦
