Problem description
| Question No.: | 202012-2 |
| Test title: | Optimal threshold of period end forecast |
| time limit: | 1.0s |
| Memory limit: | 512.0MB |
| Problem Description: | Topic backgroundConsidering that the safety index is an integer in a large range and the side dishes are likely to be confused about whether they are really safe, Dunton decided to set a threshold θ， So that the safety index y Into a specific prediction result - "fail" or "fail". Because the higher the safety index, the less likely it is for students to fail the course y≥ θ Dunton will predict that the side dishes will be very safe and will not fail this semester; Conversely, if y< θ， Dunton would admonish the side dish: "you're going to fail at the end of the term. Don't say you didn't say it beforehand." So how to set this threshold? Dunton is ready to look for answers from the past. Title DescriptionSpecifically, Dunton assessed m The safety index of students last semester, of which the second The safety index of students i (1 ≤ i ≤ m) is yi, it's a [0,108] Integer in range; At the same time, the student's failure in the last semester was recorded as resulti ∈ 0,1, where 0 Indicates failed, 1 Indicates that you have not failed. Accordingly, use immediately predict θ (y) Indicates according to the threshold θ Safety index y Translate into specific forecast results. predictθ(y)={0(y<θ)1(y≥θ) Finally, Dunton designed the following formula to calculate the optimal threshold θ ∗: θ∗=maxargmaxθ∈yi∑j=1m(predictθ(yj)==resultj) The formula can also be expressed equivalently as the following rules:
Input formatRead in data from standard input. The first line of input contains a positive integer m. Next enter m Line, where The i (1 ≤ i ≤ m) line includes two integers separated by spaces yi and resulti, as defined above. Output formatOutput to standard output. Outputs an integer representing the optimal threshold θ ∗. Example 1 input6 0 0 1 0 1 1 3 1 5 1 7 1 Data Sample 1 output3 Data Example 1 explanationAccording to rule 1, the selection range of the optimal threshold is 0,1,3,5,7. θ= 0 When, the number of correct predictions is 4； θ= one When, the number of correct predictions is 5； θ= three When, the number of correct predictions is 5； θ= five When, the number of correct predictions is 4； θ= seven When, the number of correct predictions is 3. The threshold value is selected as one or three The prediction accuracy is the highest; According to rule 3, θ ∗=max1,3=3. Example 2 Input8 5 1 5 0 5 0 2 1 3 0 4 0 100000000 1 1 0 Data Example 2 output100000000 Data Subtask70% Test data assurance m≤200； All test data assurance 2≤m≤105. |
50 points violence solution
//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<list>
#include<set>
#include<iomanip>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cassert>
#include<sstream>
#include<algorithm>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
#define ls (p<<1)
#define rs (p<<1|1)
#define mid (l+r)/2
#define over(i,s,t) for(register long long i=s;i<=t;++i)
#define lver(i,t,s) for(register long long i=t;i>=s;--i)
const int MAXN = 305;
const int INF = 0x3f3f3f3f;
const int N=5e4+7;
const int maxn=1e5+5;
const double EPS=1e-10;
const double Pi=3.1415926535897;
//inline double max(double a,double b){
// return a>b?a:b;
//}
//inline double min(double a,double b){
// return a<b?a:b;
//}
int xd[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int yd[8] = {1, 0, -1, 0, -1, 1, -1, 1};
//void Fire(){
// queue<node> p;
// p.push({fx,fy,0});
// memset(fire, -1, sizeof(fire));
// fire[fx][fy]=0;
// while(!p.empty()){
// node temp=p.front();
// p.pop();
// for(int i=0;i<8;i++){
// int x=temp.x+xd[i];
// int y=temp.y+yd[i];
// if(x<0||x>=n||y<0||y>=m||fire[x][y]!=-1){
// continue;
// }
// fire[x][y]=temp.val+1;
// p.push({x,y,temp.val+1});
// }
// }
//}
//int bfs(){
// queue<node> p;
// memset(vis, 0, sizeof(vis));
// p.push({sx,sy,0});
// while (!p.empty()) {
// node temp=p.front();
// vis[temp.x][temp.y]=1;
// p.pop();
// for(int i=0;i<4;i++){
// int x=temp.x+xd[i];
// int y=temp.y+yd[i];
// if(x<0||x>=n||y<0||y>=m) continue;
// if(x==ex&&y==ey&&temp.val+1<=fire[x][y]) return temp.val+1;
// if(vis[x][y]||temp.val+1>=fire[x][y]||a[x][y]=='#') continue;
// p.push({x,y,temp.val+1});
// }
// }
// return -1;
//}
int m;
vector<pair<int , int>>p;
vector<int > s;
int a,b;
int main(){
cin>>m;
for(int i=0;i<m;i++){
cin>>a>>b;
p.push_back({a,b});
s.push_back(a);
}
int id=-1;
int sum=0;
int maxsum=0;
for(int i=0;i<s.size();i++){
sum=0;
for(int j=0;j<p.size();j++){
if((p[j].first>=s[i]&&p[j].second==1)||(p[j].first<s[i]&&p[j].second==0))
{
sum++;
}
}
if(maxsum<=sum) {
maxsum=sum;
id=p[i].first;
}
}
cout<<id;
// for(int i=0;i<p.size();i++){
// cout<<p[i].first<<" "<<p[i].second<<endl;
// }
}
100 decomposition method
//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<list>
#include<set>
#include<iomanip>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cassert>
#include<sstream>
#include<algorithm>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
#define ls (p<<1)
#define rs (p<<1|1)
#define mid (l+r)/2
#define over(i,s,t) for(register long long i=s;i<=t;++i)
#define lver(i,t,s) for(register long long i=t;i>=s;--i)
const int MAXN = 305;
const int INF = 0x3f3f3f3f;
const int N=5e4+7;
const int maxn=1e5+5;
const double EPS=1e-10;
const double Pi=3.1415926535897;
//inline double max(double a,double b){
// return a>b?a:b;
//}
//inline double min(double a,double b){
// return a<b?a:b;
//}
int xd[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int yd[8] = {1, 0, -1, 0, -1, 1, -1, 1};
//void Fire(){
// queue<node> p;
// p.push({fx,fy,0});
// memset(fire, -1, sizeof(fire));
// fire[fx][fy]=0;
// while(!p.empty()){
// node temp=p.front();
// p.pop();
// for(int i=0;i<8;i++){
// int x=temp.x+xd[i];
// int y=temp.y+yd[i];
// if(x<0||x>=n||y<0||y>=m||fire[x][y]!=-1){
// continue;
// }
// fire[x][y]=temp.val+1;
// p.push({x,y,temp.val+1});
// }
// }
//}
//int bfs(){
// queue<node> p;
// memset(vis, 0, sizeof(vis));
// p.push({sx,sy,0});
// while (!p.empty()) {
// node temp=p.front();
// vis[temp.x][temp.y]=1;
// p.pop();
// for(int i=0;i<4;i++){
// int x=temp.x+xd[i];
// int y=temp.y+yd[i];
// if(x<0||x>=n||y<0||y>=m) continue;
// if(x==ex&&y==ey&&temp.val+1<=fire[x][y]) return temp.val+1;
// if(vis[x][y]||temp.val+1>=fire[x][y]||a[x][y]=='#') continue;
// p.push({x,y,temp.val+1});
// }
// }
// return -1;
//}
int n;
pair<int, int>p[N];
int sum[N];
set<int> s;
int main(){
cin>>n;
int a,b;
for(int i=1;i<=n;i++){
cin>>a>>b;
p[i]=make_pair(a, b);
}
sort(p+1, p+1+n);
for(int i=1;i<=n;i++)
sum[i]=sum[i-1]+p[i].second;
int ans,cnt;
int num1,num2;
for(int i=1;i<=n;i++){
int a=p[i].first;
if(s.count(a)) continue;
s.insert(a);
num1=sum[n]-sum[i-1];
num2=i-1-sum[i-1];
if((num1+num2)>=ans){
ans=num1+num2;
cnt=a;
}
}
cout<<cnt;
}