AtCoder Beginner Contest 188题解
A
题意
问\(x,y\)相差是否小于\(3\)
#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long long
ll x,y;
int main(){
scanf("%lld%lld",&x,&y);
if(abs(y - x) < 3)
puts("Yes");
else
puts("No");
}
B
题意
问两个向量的内积是否为\(0\)
照着题目模拟就行
#include<iostream>
#include<cstdio>
#define ll long long
ll n,a[10000006],b[10000006];
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i)
scanf("%lld",&a[i]);
ll sum = 0;
for(int i = 1;i <= n;++i){
scanf("%lld",&b[i]);
sum += a[i] * b[i];
}
if(sum == 0)
puts("Yes");
else
puts("No");
}
C
题意
\(2^n\)个人进行树状比赛，问最后获得第二名的人是谁
用类似于线段树的建树，最后对根节点\(1\)的两个子节点进行判断(比较的时候用键值，树上存的是编号)
#include<iostream>
#include<cstdio>
#define ll long long
#define mid ((l + r) >> 1)
ll n,a[1000005];
int val[(1000005) << 2];
void build(int now,int l,int r){
if(l == r){
val[now] = l;
return ;
}
build(now * 2,l,mid);
build(now * 2 + 1,mid + 1,r);
if(a[val[now * 2]] > a[val[now * 2 + 1]])
val[now] = val[now * 2];
else
val[now] = val[now * 2 + 1];
return;
}
int main(){
scanf("%lld",&n);
n = (1 << n);
for(int i = 1;i <= n;++i)
scanf("%lld",&a[i]);
build(1,1,n);
if(val[1] == val[2])
std::cout<<val[3];
else
std::cout<<val[2];
}
D
题意
\(Takahashi\) 出去游玩，现在提供了\(n\)个项目，时间是\([a_i, b_i]\)，这些项目每天分别需要花费\(c_i\)，但它可以选择一天花费\(C\)元玩这天所有可玩项目，项目出现了他就一定要玩，求他的最少花费
考虑离散化，在离散后的数组上打差分标记，再对当前的钱数进行一个前缀和，对每一段判断是否取\(C\)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
ll n,C;
ll a[200005],b[200005],c[200005],cnt;
ll num[800005],mark[800005];
unsigned ll now = 0,ans = 0;
int main(){
scanf("%lld%lld",&n,&C);
ll sum = 0;
ll vsum = 0;
for(int i = 1;i <= n;++i){
scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);
num[++cnt] = a[i];
num[++cnt] = b[i] + 1;
}
std::sort(num + 1,num + cnt + 1);
cnt = std::unique(num + 1,num + cnt + 1) - num - 1;
for(int i = 1;i <= n;++i){
mark[std::lower_bound(num + 1,num + cnt + 1,a[i]) - num] += c[i];
mark[std::lower_bound(num + 1,num + cnt + 1,b[i] + 1) - num] -= c[i];
}
num[0] = num[1] + 1;
// for(int i = 1;i <= cnt;++i)
// std::cout<<num[i]<<" "<<mark[i]<<std::endl;
now = mark[1];
for(int i = 2;i <= cnt;++i){
//std::cout<<" "<<num[i - 1]<<" "<<num[i]<<" "<<now<<std::endl;
if(now <= C)
ans += (now) * (num[i] - num[i - 1]);
else
ans += C * (num[i] - num[i - 1]);
now += mark[i];
}
std::cout<<ans<<std::endl;
}
E
题意
有向图，有点权，你可以在一个点上用点权买一块黄金，在再另一块你能到达的点上以点权卖出，问必须进行一次买卖的最大收益
我原本是进行一次\(dfs\)这样处理每个点的前缀最小值，但这样会在环上跑\(2\)次，我也这样\(T\)了一发
我们可以考虑按照点权的大小优先\(dfs\)这样最算有环也只用跑\(1\)次
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
ll n,m,cnt,ans = -0x3f3f3f3f;
ll head[200005];
ll val[200005],maxx[200005],minn[200005];
int vis[200005];
struct K{
ll val,num;
}p[200005];
struct P{
ll to,next;
}e[400005];
void add(ll x,ll y){
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
}
void dfs(ll now){
//std::cout<<now<<" "<<val[now]<<" "<<minn[now]<<std::endl;
ans = std::max(val[now] - minn[now],ans);
for(int i = head[now];i;i = e[i].next){
if(!vis[e[i].to]){
minn[e[i].to] = std::min(minn[now],val[now]);
vis[e[i].to] ++ ;
dfs(e[i].to);
}
}
}
bool operator < (K a,K b){
return a.val < b.val;
}
int main(){
memset(maxx,-0x3f,sizeof(maxx));
memset(minn,0x3f,sizeof(minn));
scanf("%lld%lld",&n,&m);
for(int i = 1;i <= n;++i)
scanf("%lld",&p[i].val),p[i].num = i,val[i] = p[i].val;
for(int i = 1;i <= m;++i){
ll x,y;
scanf("%lld%lld",&x,&y);
add(x,y);
}
std::sort(p + 1,p + 1 + n);
for(int i = 1; i<= n;++i)
if(!vis[p[i].num])
dfs(p[i].num);
std::cout<<ans<<std::endl;
}
F
题意
有\(x,y\),可以对\(x做+1,-1,*2\)的操作请问最少几次能到\(y\)
考虑进行记忆化搜索，以后这种没有太好思路的题都可以往搜索想
#include<iostream>
#include<cstdio>
#include<map>
#define ll long long
using std::map;
ll x,y;
map<ll,ll>QWQ;
ll solve(ll y){
if(y <= x) return x - y;
if(QWQ.count(y)) return QWQ[y];
ll ans = y - x;
if(y % 2) ans = std::min(ans,1 + std::min(solve(y - 1),solve(y + 1)));
else
ans = std::min(ans,1 + solve(y / 2));
return QWQ[y] = ans;
}
int main(){
scanf("%lld%lld",&x,&y);
std::cout<<solve(y);
}
