题解分享
题解分享简介
卷王 - 题解
include
using namespace std;
int main(){
int n;
cin>>n;
int a[n],b[n],c[n];
for(int i=0;i
>c[i];
}
for(int i=0;i
>a[i];
}
for(int i=0;i
>b[i];
}
int sum_a=0,sum_b=0;
for(int i=0;i
sum_b){
cout<<"ke"<<endl;
}
else if(sum_a<sum_b){
cout<<"do"<<endl;
}
else{
cout<<"same"<<endl;
}
return 0;
}
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卷王 - 题解
```
/*
思路分析 ：
1.三个数组接收三组数据
*/
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin >> n;
vector<double> a;
vector<double> b;
vector<double> c;
for ( int i = 0; i < n; i++ ) {
double temp;
cin >> temp;
a.push_back(temp);
}
for ( int i = 0; i < n; i++ ) {
double temp;
cin >> temp;
b.push_back(temp);
}
for ( int i = 0; i < n; i++ ) {
double temp;
cin >> temp;
c.push_back(temp);
}
double tuoTuo = 0;
double xiaoMing = 0;
for ( int j = 0; j < n; j++ ) {
tuoTuo += a[j] * b[j] / 100;
xiaoMing += a[j] * c[j] /100;
}
//cout << tuoTuo << " " << xiaoMing;
// 因为是百分率 所以保留两位小数进行比较
tuoTuo = round(tuoTuo * 100) / 100;
xiaoMing = round(xiaoMing * 100) / 100;
if ( tuoTuo < xiaoMing ) {
cout << "do";
}
else if ( tuoTuo > xiaoMing ) {
cout << "ke";
}
else {
cout << "same";
}
return 0;
}
```
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卷王 - 题解
```
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
cin >> N;
vector<int> score(N);
for(int i=0; i<N; i++)
cin >> score[i];
vector<int> sum(2,0);
int rate;
for(int i=0; i<2; i++)
for(int j=0; j<N; j++)
{
cin >> rate;
sum[i]+=score[j]*rate;//题目数据给的太仁慈了
}
if(sum[0]>sum[1])cout << "ke" << endl;
else if (sum[0]<sum[1])cout << "do" << endl;
else cout << "same" << endl;
return 0;
}
```
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卷王 - 题解
```
#include<cstdio>
#include<iostream>
using namespace std;
const int N=1e5+0;
int a[N];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n,num;
cin>>n;
long long sumb=0,sumc=0;
for(int j=0;j<n;++j) cin>>a[j];
for(int j=0;j<n;++j){
cin>>num;
sumb+=a[j]*num;
}
for(int j=0;j<n;++j){
cin>>num;
sumc+=a[j]*num;
}
if(sumb>sumc) cout<<"ke";
else if(sumb==sumc) cout<<"same";
else cout<<"do";
return 0;
}
```
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卷王 - 题解
```
#include<bits/stdc++.h>
using namespace std;
double tt_total=0,xm_total=0;
int n;
vector<int >score;
vector<double>tt_p;
vector<double>xm_p;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
double x;
cin>>x;
score.push_back(x);
}
for(int i=0;i<n;i++)
{
double x;
cin>>x;
tt_p.push_back(x);
}
for(int i=0;i<n;i++)
{
double x;
cin>>x;
xm_p.push_back(x);
}
for(int i=0;i<n;i++)
{
tt_total+=score[i]*1.0*(tt_p[i]*1.0/100);
xm_total+=score[i]*1.0*(xm_p[i]*1.0/100);
}
double eps=1e-6;//精度默认定义为1e-6，当fabs（x-y）<eps时，认为x==y。
if(fabs(tt_total-xm_total)<eps)cout<<"same";//fabs（x）：计算x的绝对值
else if(tt_total>xm_total)cout<<"ke";
else cout<<"do";
return 0;
}
```
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卷王 - 题解
```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[3][n];
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < n; ++j) {
cin >> a[i][j];
}
}
int tt = 0, xm = 0;
for (int i = 0; i < n; ++i) {
tt += a[0][i] * a[1][i];
xm += a[0][i] * a[2][i];
}
if (tt > xm) {
cout << "ke";
} else if (tt == xm) {
cout << "same";
} else {
cout << "do";
}
}
```
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