Rounded small trick

Question meaning: Portal
Solution: If the classification is based on $codeforces$ is the simplest$implemention$ , but it stuck me for a long time, and the meaning of the question alsoemphasizes that(a bitter face) needs to be rounded, so what are the small tricks for rounding? First understand that there is a$R\; \&\; ltOUn-D$ function, it may be directly used, thisarticleso that nothing in fact have to be programmed in terms of tangled, such as another is directly$x\; is$ rounded to a whole number, thenx=(x*1+0.5)/1.0, then rounding to a few decimal places is also written,x=(x*100+0.5)/100.0that is,to keep two decimal places.
Attach the code (use the round function directly):

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define rush() int T;scanf("%d",&T);while(T--)
#define mm(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define fi first
#define se second
#define db double
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=70;
db stus[N][10];
int n;
db sum;
int main()
{
    
    
    sc(n);
    rep(i,1,n){
    
    
        rep(j,1,8){
    
    
            db t;
            scanf("%lf",&t);
            if(j==2&&stus[i][1]!=2)sum+=t;
            stus[i][j]=t;
        }
        if(stus[i][1]==2)stus[i][9]=0;
        else{
    
    
            stus[i][9]+=round(stus[i][3]*stus[i][4]+stus[i][5]*stus[i][6]+stus[i][7]*stus[i][8]);
        }
    }
    db res=0;
    rep(i,1,n){
    
    res+=stus[i][9]*(stus[i][2]/sum);}
    printf("%.2f\n",res);
    return 0;
}

The second type: round up manually by yourself

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define rush() int T;scanf("%d",&T);while(T--)
#define mm(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define fi first
#define se second
#define db double
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=70;
db stus[N][10];
int n;
db sum;
int main()
{
    
    
    sc(n);
    rep(i,1,n){
    
    
        rep(j,1,8){
    
    
            db t;
            scanf("%lf",&t);
            if(j==2&&stus[i][1]!=2)sum+=t;
            stus[i][j]=t;
        }
        if(stus[i][1]==2)stus[i][9]=0;
        else{
    
    
            for(int k=3;k<=7;k+=2){
    
    
                stus[i][9]+=stus[i][k]*stus[i][k+1];
            }
        }
    }
    db res=0;
    rep(i,1,n){
    
    
        int g=(int)(stus[i][9]*1+0.5)/1.0;
        res+=(g*(stus[i][2]/sum));
    }
    res=(int)(res*100+0.5)/100.0;
    printf("%.2f\n",res);
    return 0;
}