Title Description
Small black to run an art school, because the last year to run the class response is good, so he expanded to 3 classes this year, he wanted to know if he did this class is more attractive to girls or boys more attractive, of course, he also want to know how many people recruit! Come on, help the black bar, you need to calculate the proportion of female students, the proportion of boys, including the number of classes, the whole school, as well as the total numbers of classes and the total number of recruits.
Input
A total of three classes, respectively, the number of boys to enter the class, girls
Output
First output the number of classes, and then output each class of boys accounted for the total number of students, girls accounted for the total number of the proportion. Finally, the total number of students, and the proportion of boys accounted for the total number of the school, girls accounted for the total number of the school.
When the scale is output, one decimal place is reserved and the total is not used.
Sample input
13 2114 2010 22
Sample output
34 0.4 0.634 0.4 0.632 0.3 0.7100 0.4 0.6
The code is as follows:
#include <iostream> #include <iomanip>using namespace Std;int main () { int a[3][2],sum[4]={0,0,0,0},m= 0,n=0,i,j; Double b[4][2]; for (i=0;i<3;i++) {for (j=0;j<2;j++) { cin>>a[i][j]; } } for (i=0;i<3;i++) {for (j=0;j<2;j++) { sum[i]+=a[i][j]; } Sum[3]+=sum[i]; M+=A[I][0]; N+=A[I][1]; } for (i=0;i<3;i++) {for (j=0;j<2;j++) { b[i][j]=a[i][j]*1.0/sum[i]; } } for (i=0; i<3; i++) { cout<<sum[i]<< ""; Cout<<setiosflags (ios::fixed) <<setprecision (1); cout<<b[i][0]<< "" <<b[i][1]<<endl; } cout<<sum[3]<< ""; Cout<<setiosflags (ios::fixed) <<setprecision (1); cout<<m*1.0/sum[3]<< "" <<n*1.0/sum[3]<<endl; return 0;}
Operation Result:
This topic actually stuck me for more than half an hour, sure enough data a lot is easy to divide.
OJ Brush question "Calculate the proportion of boys and girls"