202012 CCF test questions

202012-1 Safety index of final forecast

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

// #define debug

int main(){
    
    #ifdef debug
        freopen("test.in", "r", stdin); 
    #endif
    int n,w,s;
    cin>>n;
    ll sum=0;
    for(int i=0;i<n;++i){
    
        cin>>w>>s;
        sum += w*s;
        // cout<<sum<<endl;
    }
    cout << ((sum > 0)?sum:0);
}

202012-2 The best threshold of final forecast

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+5 ; 

// #define debug

struct Predict{
    
    int y;
    int r;
    int n0; //  Statistics 0 The number of 
    int n1; //  Statistics 1 The number of 
    Predict(int y=0,int r=0,int n0=0,int n1=0):y(y),r(r),n0(n0),n1(n1){
    }
    void setn(int n0,int n1){
    
        this->n0 = n0;
        this->n1 = n1;
    }
};


bool cmp(const Predict& a, const Predict& b)
{
    
	return a.y < b.y; // Sort from large to small 
}

Predict pred[MAXN];

/*  According to the general meaning of the topic , First count each number （ It's going to repeat itself ） Marked as 0,1 The number of , Then give a number , Find the sign less than this number 0 The number of and greater than or equal to are marked as 1 Number of numbers   It is equivalent to dividing into two sections , Left side , Cumulative value on the right , For accumulation, we can use prefix and to do   so ： Prioritize , duplicate removal , Find the prefix and , Find the result  */
int main(){
    
    #ifdef debug
        // freopen("test.in", "r", stdin); // 3
        freopen("test1.in", "r", stdin); // 100000000
        // freopen("test.out","w", stdout);
    #endif
    int m,tmp=1,result = 0,maxr=0;
    cin>>m;
    for(int i=1;i<=m;++i){
    
        cin>>pred[i].y>>pred[i].r;
        if(pred[i].r==1) {
    
            pred[i].setn(0,1);
        } else {
    
            pred[i].setn(1,0);
        }
    }
    // m++;
    sort(pred+1,pred+m+1,cmp);
     //  duplicate removal 
    for(int i=2;i<=m;++i) {
    
        if(pred[i].y==pred[i-1].y){
    
            pred[tmp].n1 += pred[i].n1;
            pred[tmp].n0 += pred[i].n0;
        } else {
    
            pred[++tmp].y =  pred[i].y;
            pred[tmp].n1 = pred[i].n1;
            pred[tmp].n0 = pred[i].n0;
        }
    }
    m = tmp;
    for(int i=1;i<=m;++i){
    
        pred[i].n0 += pred[i-1].n0;
        pred[i].n1 += pred[i-1].n1;
    }
    for (int i=1;i<=m;++i){
    
        tmp = pred[i-1].n0 + pred[m].n1 - pred[i-1].n1; //  Duplicate data 
        // cout<<tmp<<endl;
        if (tmp>=maxr) {
    
            maxr = tmp;
            result = pred[i].y;
        }
    }
    cout<<result;
}