The ninth provincial competition
text
Spiral Polyline
Topic description
对于整点(X, Y)，我们定义它到原点的距离dis(X, Y)是从原点到(X, Y)的螺旋折线段的长度。
例如dis(0, 1)=3, dis(-2, -1)=9
给出整点坐标(X, Y)，你能计算出dis(X, Y)吗？
输入格式
X和Y
对于40%的数据，-1000 <= X, Y <= 1000
对于70%的数据，-100000 <= X， Y <= 100000
对于100%的数据, -1000000000 <= X, Y <= 1000000000
输出格式
输出dis(X, Y)
样例输入
0 1
样例输出
3
Topic ideas and codes
Idea: As long as the coordinate law of all points on the Y-axis is found (called the horizontal key point), that is, the distance to the same horizontal line where the point is located can also be found.
当y > 0 时：
当abs(x) <= y时，dis(0 , y）=3 * y + (y * y - y) / 2 * 8，所以dis(x , y)=dis(0 , y) + x;
当abs(x) > 时, x > 0时，dis(x , y) = dis(0 , x) + 2 * x - y。
x < 0时，dis(x , y) = dis(0 ,-x) + 2 * x + y。
当y <= 0时：
当y-1 <= x <= -y 时，dis(0 , -y) = 7 * -y + (y * y + y)/2 * 8,所以dis(x , y) =dis(0 , y) - x;
当x >- y 或 x< y - 1时，x > 0 时，dis(x,y) = dis(0 , x) - 2 * x - y。
x < 0 时，dis(x,y) = dis(0 , -x - 1) - 2 * x + y - 1。
Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> P;
const ll maxn=1e3+10;
ll x,y;
int main(){
cin>>x>>y;
ll n=max(abs(x),abs(y));
ll dx=-n,dy=-n;
ll ans=0;
ans=n*(n-1)*4;
if(y==-n){
ans+=3*n*2+n-x;
}
if(y==n){
ans+=2*n+x+n;
}
if(x==-n){
ans+=n+y;
}
if(x==n){
ans+=n*2*2+n-y;
}
cout<<ans;
return 0;
}
maximum product
Topic description
给定N个整数A1, A2, ... AN。请你从中选出K个数，使其乘积最大。
请你求出最大的乘积，由于乘积可能超出整型范围，你只需输出乘积除以1000000009的余数。
注意，如果X<0， 我们定义X除以1000000009的余数是负(-X)除以1000000009的余数。
即：0-((0-x) % 1000000009)
输入格式
第一行包含两个整数N和K。
以下N行每行一个整数Ai。
对于40%的数据，1 <= K <= N <= 100
对于60%的数据，1 <= K <= 1000
对于100%的数据，1 <= K <= N <= 100000 -100000 <= Ai <= 100000
输出格式
一个整数，表示答案。
样例输入
5 3
-100000
-10000
2
100000
10000
样例输出
999100009
再例如：
样例输入
5 3
-100000
-100000
-2
-100000
-100000
样例输出
-999999829
Topic ideas and codes
First sort according to the absolute value from largest to smallest
①If there must be 0 in the selected k numbers, the result is 0
②If they are all negative numbers, then if k is an even number, you can select the first k numbers, if k is Odd numbers, you can only select k numbers from the small start.
③If they are all positive numbers, you can select the first k numbers.
④If there are both positive and negative numbers, then
- If the first k numbers have an even number of negative numbers, select the first k numbers
- If the first k numbers have an odd number of negative numbers, then we can see if we can choose a large positive number from the back to exchange with the small negative number in front, or choose a large negative number from the back to exchange with the small positive number in front, and take the result of the two. Big.
At this point, if the result is still negative after all cases are divided, we check whether the input has 0, and the result is 0 if there is zero, otherwise, only this negative number can be output.
code
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> P;
const ll maxn=1e5+10;
const ll mod=1000000009;
ll n;
ll a[maxn];
ll k;
int main(){
cin>>n>>k;
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
sort(a+1,a+1+n);
if(n==k){
ll res=1;
for(int i=1;i<=n;i++){
res=res*a[i]%mod;
}
cout<<res;
return 0;
}
if(a[n]<0){
if(k&1){
ll res=1;
for(int i=n-k+1;i<=n;i++){
res=res*a[i]%mod;
}
cout<<res<<endl;
}
else{
ll res=1;
for(int i=1;i<=k;i++){
res=res*a[i]%mod;
}
cout<<res<<endl;
}
return 0;
}
ll l=1,r=n;
ll res=1;
if(k&1){
res=a[r];
r--;
k--;
//if(res<0)sign=-1;
}
while(k){
ll x=(ll)a[r]*a[r-1],y=(ll)a[l]*a[l+1];
if(x>=y){
r-=2;
res=x%mod*res%mod;//先对x进行取模，在对乘积结果进行取模,原因是x也是乘积结果如果不取模乘上的结果不对
}
else{
l+=2;
res=y%mod*res%mod;//先对y进行取模，在对乘积结果进行取模
}
k-=2;
}
cout<<res;
return 0;
}
Incrementing triples
Topic description
给定三个整数数组
A = [A1, A2, ... AN],
B = [B1, B2, ... BN],
C = [C1, C2, ... CN]，
请你统计有多少个三元组(i, j, k) 满足：
1. 1 <= i, j, k <= N
2. Ai < Bj < Ck
输入格式
第一行包含一个整数N。
第二行包含N个整数A1, A2, ... AN。
第三行包含N个整数B1, B2, ... BN。
第四行包含N个整数C1, C2, ... CN。
对于30%的数据，1 <= N <= 100
对于60%的数据，1 <= N <= 1000
对于100%的数据，1 <= N <= 100000 0 <= Ai, Bi, Ci <= 100000
输出格式
一个整数表示答案
样例输入
3
1 1 1
2 2 2
3 3 3
样例输出
27
Topic ideas and codes
Idea: You can sort the three arrays first, then traverse the array b to find how many of the a array are smaller than b[i] and how many of the c array are greater than b[i]
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> P;
const ll maxn=1e5+10;
const ll mod=1000000009;
ll n;
ll a[maxn],b[maxn],c[maxn];
int main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
}
for(int i=1;i<=n;i++){
cin>>b[i];
}
for(int i=1;i<=n;i++){
cin>>c[i];
}
sort(a+1,a+1+n);
sort(b+1,b+1+n);
sort(c+1,c+1+n);
ll sum=0;
for(int i=1;i<=n;i++){
ll x=lower_bound(a+1,a+1+n,b[i])-a;
ll y=upper_bound(c+1,c+1+n,b[i])-c;
sum=sum+(x-1)*(n-y+1);
//cout<<x<<" "<<y<<endl;
}
cout<<sum;
return 0;
}
Epilogue
"If you are undecided, you can ask the spring breeze, and if the spring breeze does not speak, you will follow your heart" means: if you are hesitant about something, ask the spring breeze how to do it. . "If you are undecided, you can ask the spring breeze. If the spring breeze does not speak, you will follow your heart." The sentence comes from the "Jianlai" written by the Internet writer "Fenghuo Opera Princes". The original text is: "If you are undecided, you can ask the spring breeze. Follow your heart".
