2019 杭电多校（第六场）
1005 Snowy Smile （线段树）
http://acm.hdu.edu.cn/showproblem.php?pid=6638
题意
给你n个点 让你画个矩形 使矩形内所含点的权值和最大（必须有点）
思路
离散化 枚举矩形的左右区间 线段树维护y坐标的最大字段和 （复杂度 O(n*n*lgn))
代码
#include
using namespace std;
typedef long long ll;
const int maxn = 2e3+100;
int hax[maxn],hay[maxn];
struct point
{
int x,y;
ll w;
}po[maxn];
struct node
{
ll sum,lmax,rmax,MAX;
}tree[maxn*4];
int n;
bool cmp(point a,point b)
{
if(a.x == b.x) return a.y < b.y;
else return a.x < b.x;
}
void build(int x,int l,int r)
{
tree[x].sum = tree[x].lmax = tree[x].rmax = tree[x].MAX = 0;
if(l == r)
{
return ;
}
int mid = (l + r) / 2;
build(x*2,l,mid);
build(x*2+1,mid+1,r);
}
void updata(int x,int l,int r,int id,ll w)
{
if(l==r)
{
tree[x].sum += w;
tree[x].lmax += w;
tree[x].rmax += w;
tree[x].MAX += w;
return ;
}
int mid = (l + r) / 2;
if(id <= mid) updata(x*2,l,mid,id,w);
else updata(x*2+1,mid+1,r,id,w);
tree[x].sum = tree[x*2].sum + tree[x*2+1].sum;
tree[x].lmax = max(tree[x*2].lmax,tree[x*2].sum + tree[x*2+1].lmax);
tree[x].rmax = max(tree[x*2+1].rmax, tree[x*2+1].sum + tree[x*2].rmax);
tree[x].MAX = max(max(tree[x*2].MAX,tree[x*2+1].MAX),tree[x*2].rmax + tree[x*2+1].lmax);
}
ll query()
{
return tree[1].MAX;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = 1;i <= n;i++)
{
scanf("%d%d%lld",&po[i].x,&po[i].y,&po[i].w);
hax[i] = po[i].x,hay[i] = po[i].y;
}
sort(hay+1,hay+1+n);
int mm = unique(hay+1,hay+1+n) - (hay+1);
sort(po+1,po+1+n,cmp);
ll ans = 0;
po[n+1].x = 13452353;
for(int i = 1;i <= n;)
{
build(1,1,mm);
for(int j = i;j <= n;j++)
{
int id = lower_bound(hay+1,hay+1+mm,po[j].y) - hay;
updata(1,1,mm,id,po[j].w);
if(po[j].x != po[j+1].x)
{
ll sum = query();
ans = max(ans,sum);
}
}
i++;
while(po[i].x == po[i-1].x)
{
i++;
}
}
printf("%lld\n",ans);
}
return 0;
}
10006 Faraway
http://acm.hdu.edu.cn/showproblem.php?pid=6639
题意
求解方程
思路
对于每个式子 去绝对值后分为四个区间（加一横一竖）n个式子 就是n*n个区间
对着n*n个区间求解 将数分成 lcm(2,3,4,5) = 60种 即如果(x,y)满足这n个式子 那么(x+60,y+60)也一定满足这个式子
对每个区间60*60验证所有类型的数的组合 然后验证n个式子是否满足 满足的话计算着个区间里有多少种答案
代码
#include
using namespace std;
typedef long long ll;
const int maxn = 20;
struct node
{
ll x,y,k,t;
}a[maxn];
ll hax[maxn],hay[maxn];
ll n;
int check(int x,int y)
{
for(int i = 1;i <= n;i++)
{
if((abs(x-a[i].x) + abs(y-a[i].y)) % a[i].k != a[i].t) return 0;
}
return 1;
}
ll ok(ll l,ll r)
{
ll len = r - l - 1;
if(len < 0) return 0;
return len / 60 + 1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
ll m;
scanf("%lld%lld",&n,&m);
hax[1] = hay[1] = m+1;
for(ll i = 1;i <= n;i++)
{
scanf("%lld%lld%lld%lld",&a[i].x,&a[i].y,&a[i].k,&a[i].t);
hax[i+1] = a[i].x,hay[i+1] = a[i].y;
}
sort(hax+1,hax+1+n+1);
sort(hay+1,hay+1+n+1);
ll ans = 0;
for(int i = 0;i < n+1;i++)
{
if(hax[i] == hax[i+1]) continue;
for(int j = 0;j < n+1;j++)
{
if(hay[j] == hay[j+1]) continue;
// cout< 1008 TDL
http://acm.hdu.edu.cn/showproblem.php?pid=6641
题意
f(n,m) 表示大于n的第m小的与n互斥的数 (f(n,m)−n)^n=k 给你k m 求最小n
思路
m小于100 那么f(n,m)最大不会超过n+1000
设f(n,m) 为n + sum
(n + sum - n) ^ n == k ==> sum = k^n ==> n = sum^k
枚举sum从1到1000 求出n排序 挨个验证
代码
#include
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
ll a[2200];
int ok(ll n,ll m,ll k)
{
ll x = 1;
for(ll i = n + 1;;i++)
{
if((__gcd(n,i)) == 1)
{
if(x == m)
{
ll aa = (i-n)^n;
if(aa == k)
{
return 1;
}
else return 0;
}
x++;
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
ll m,k,n;
scanf("%lld%lld",&k,&m);
int f = 0;
for(ll i = 1;i <= 2000;i++)
{
a[i] = k ^ i;
}
sort(a+1,a+1+2000);
for(ll i = 1;i <= 2000;i++)
{
n = a[i];
if(n == 0) continue;
if(ok(n,m,k))
{
printf("%lld\n",n);
f = 1;
break;
}
}
if(f == 0) printf("-1\n");
}
return 0;
} 1012 Stay Real
题意
。。。。。。
思路
签到题
代码
#include
using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
int a[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
ll num = 0;
for(int i = 1;i <= n;i++)
{
scanf("%lld",&a[i]);
num += a[i];
}
ll sum = 0;
sort(a+1,a+1+n);
for(int i = n;i >= 1;i-=2)
{
sum += a[i];
}
printf("%lld %lld\n",sum,num-sum);
}
return 0;
}