hdu 多校賽 第八場

首先確定答案不會太大，寫個狀壓跑一跑，能跑出所有答案（或者直接oeis）

#include #define ll long long

#define sc scanf
#define pr printf
using namespace std;
ll a1[84] = ;
ll a2[14] = ;
ll a3[36] = ;
ll a4[48] = ;
ll a5[5] = ;
ll a6[72] = ;
ll a7[49] = ;
ll a8[344] = ;
ll a9[9] = ;
int main()}}
else if (d == 2)}}
else if (d == 3)}}
else if (d == 4)}}
else if (d == 5)}}
else if (d == 6)}}
else if (d == 7)}}
else if (d == 8)}}
else if (d == 9)}}
qwe:;
}}

先離散化，然後暴力每條邊會產生的不連通區域，然後跑dfs連通塊。

#include #define ll long long

#define sc scanf
#define pr printf
using namespace std;
struct point
}a[4], b[4];
ll x[10];
ll y[10];
int s[30][30];
bool vis[30];
void calc(point q, point w)
else if (w.y == 2)
else if (w.y == 3)
else if (w.y == 4)
}else if (q.y == 2)
else if (w.y == 3)
else if (w.y == 4)
}else if (q.y == 3)
else if (w.y == 4)}}
else
else if (w.x == 2)
else if (w.x == 3)
else if (w.x == 4)
}else if (q.x == 2)
else if (w.x == 3)
else if (w.x == 4)
}else if (q.x == 3)
else if (w.x == 4)}}
}void dfs(int index) 
}}int main()
for (int i = 1; i <= 25; i++)
if (i % 5 != 1)
}/*for (int i = 1; i <= 25; i++)
for (int j = 1; j <= 25; j++)
printf("%d%c", s[i][j], j == 25 ? '\n' : ' ');*/
sc("%lld%lld%lld%lld", &a[0].x, &a[0].y, &a[2].x, &a[2].y);
sc("%lld%lld%lld%lld", &b[0].x, &b[0].y, &b[2].x, &b[2].y);
x[0] = a[0].x; x[1] = a[2].x; x[2] = b[0].x; x[3] = b[2].x;
y[0] = a[0].y; y[1] = a[2].y; y[2] = b[0].y; y[3] = b[2].y;
sort(x, x + 4);
a[0].x = lower_bound(x, x + 4, a[0].x) - x + 1;
a[2].x = lower_bound(x, x + 4, a[2].x) - x + 1;
b[0].x = lower_bound(x, x + 4, b[0].x) - x + 1;
b[2].x = lower_bound(x, x + 4, b[2].x) - x + 1;
sort(y, y + 4);
a[0].y = lower_bound(y, y + 4, a[0].y) - y + 1;
a[2].y = lower_bound(y, y + 4, a[2].y) - y + 1;
b[0].y = lower_bound(y, y + 4, b[0].y) - y + 1;
b[2].y = lower_bound(y, y + 4, b[2].y) - y + 1;
//        a[0].print();
//        a[2].print();
//        b[0].print();
//        b[2].print();
a[1] = point;
a[3] = point;
b[1] = point;
b[3] = point;
for (int i = 0; i < 4; i++)
/*for (int i = 1; i <= 25; i++)
for (int j = 1; j <= 25; j++)
printf("%d%c", s[i][j], j == 25 ? '\n' : ' ');*/
memset(vis, false, sizeof(vis));
int color = 0;
for (int i = 1; i <= 25; ++i) 
}printf("%d\n", color);
}}

#include #define ll long long

#define sc scanf
#define pr printf
#define lson left,mid,k<<1
#define rson mid+1,right,k<<1|1
#define imid int mid=(left+right)/2;
using namespace std;
const int maxn = 1e5 + 5;
struct node
que[100005];
int main()
);        ll qq = (ll)n * d;
if (qq % 10 != 5)
pr("quailty is very great\n");
else
}}

貪心，列舉班，這個班不喝自己的茶的能喝的最大的。

#includeusing namespace std;

#define ll long long
#define sc scanf
#define pr printf
const int n =1e6 + 10;
ll a[n], b[n];
int main()
ll ans = 0;
for (int i = 1; i <= n; i++)
printf("%lld\n", ans);
}}

多校第八場

hdu 4950 monster 小模擬題，找到判斷的臨界條件 include include includeusing namespace std int64 h,a,b,k,sum int main int i 0 while scanf i64d i64d i64d i64d h,a,b,k ...

多校第八場

hdu 4950 monster 小模擬題。找到推斷的臨界條件。include include includeusing namespace std int64 h,a,b,k,sum int main int i 0 while scanf i64d i64d i64d i64d h,a,b,k ...

2019 牛客 多校賽 第八場

乙個區間的貢獻為這個區間的不同數字的個數，求所有子區間的貢獻和 反向思考，計算乙個數字在子區間 現的次數，當區間的左端點在這個數字上一次出現位置和當前位置之間，並且右端點在當前位置之後，這個數字都產生了貢獻，所以當前位置的數字的貢獻就是左端點數量 右端點數量，然後遍歷一遍。include defin...