洛古最簡單50題解(21-30)

作爲一名新手,首先要過一過題,找找成就感。(大佬略過)。下面附上洛古最簡單50題(大佬略過)。以及最麻煩 AC代碼,至少AC了ios

NO.21 P1421 小玉買文具 c++

#include<iostream>
using namespace std;
int main()
{
    int a,b,x;
    cin>>a>>b;
    x=(a*10+b)/19;
    cout<<x<<endl;
    return 0;
}
spa

 

NO.22 P1422 小玉家的電費 遊戲

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
    float b,a;
cin>>a;
    if (a<=150)
    {
        b=a*0.4463;
    }
    else
    if (a>=151 && a<=400)
    {
        b=150*0.4463+(a-150)*0.4663;
    }
    else
    {
        b=150*0.4463+250*0.4663+(a-400)*0.5663;
    }
    printf("%0.1f",b);
    return 0;
}
ci

 

NO.23 P1423 小玉在游泳 it

#include<iostream>
using namespace std;
int main()
{
    int n;
    double a,s,d;
    n=0;
    s=2;
    d=0;
    cin>>a;
    while(d<a)
    {
        d=s+d;
        s=s*0.98;
        n++;
    }
    cout<<n<<endl;
    return 0;
}
io

 

NO.24 P1424 小魚的航程(改進版) stream

#include<bits/stdc++.h>
using namespace std;
int main()
{
unsigned long long n,ans=0;
int x;
cin >> x >> n;
for(int i=0;i<n;i++)
{
if((x!=6)&&(x!=7))
ans += 250;
if(x==7)
x=1;
else
x++;
}
cout << ans;
return 0;
}
float

 

NO.25 P1425 小魚的游泳時間 統計

#include<iostream>
using namespace std;
int main()
{
    int a,b,a1,b1,a2,b2;
    cin>>a>>b>>a1>>b1;
    if (b1>b)
    {
        b2=b1-b;
        a2=a1-a;
    }
    else
    {
        b2=b1+60-b;
        a2=a1-1-a;
    }
    cout<<a2<<" "<<b2<<endl;
}

 

NO.26 P1427 小魚的數字遊戲

#include<bits/stdc++.h>
using namespace std;
int a[101],n=0;
int main()
{
    a[0]=-1;
    while (a[n]!=0)
    {
        n++;
        cin>>a[n];
    }
    while(n!=1)
    {
        n--;
        cout<<a[n]<<" ";
        
    }
    return 0;
}

 

NO.27 P1428 小魚比可愛

#include<iostream>
using namespace std;
int main()
{
    int n,a[100001],b[100001];
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        b[i]=0;
    }
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        for(int j=i;j>=1;j--)
        {
            if (a[i]>a[j])
            {
                b[i]++;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        cout<<b[i]<<" ";
    }
    return 0;
}

 

NO.28 P1553 數字反轉(升級版)

#include <iostream>
using namespace std;
int main()
{
char s[30]={},t;
cin>>s;
int i=0,j,flag=0;
for (j=0; s[j]!='\0'; j++) {
t=s[j+1];
if (t=='.'||t=='/'||t=='%'||t=='\0') {
while (flag &&s[i]=='0') i++;
unsigned long long n=s[j]-'0';
for(int a=i,b=j;a<b;a++,b--){
char c=s[a];s[a]=s[b];s[b]=c;
}
i=++j+1;
if (t=='.') {
flag=1;
}
}
}
for(flag=1,j=0; s[j]!='\0'; j++){
t=s[j+1];
if(flag && s[j]=='0'){
continue;
}
if(flag && (s[j]=='.'||s[j]=='/'||s[j]=='%')){
cout<<'0';
}
flag=0;
cout<<s[j];
if (s[j]=='.'||s[j]=='/') {
flag=1;
}
}
if(flag){
cout<<'0';
}
return 0;
}

 

NO.29 1554 夢中的統計

#include<iostream>
using namespace std;
int n,m,a[10]={0};
int jk(int n,int i)
{
    int t,ans=0;
        t=i;
        while (t!=0)
        {
            if (t % 10 == n)
            {
                ans++;
            }
                t/=10;
        }
    return ans;
}
int main()
{
    cin>>n>>m;
    for(int i=n;i<=m;i++)
    {
        for(int j=0;j<=9;j++)
            a[j]+=jk(j,i);
    }
    for(int i=0;i<=9;i++)
    cout<<a[i]<<" ";
}

 

NO.30 P1567 統計天數

#include<iostream>
using namespace std;
int main()
{
    int n,a[100000000],max,t;
    cin>>n;
    max=0;
    t=0;
    for(int i=1;i<n;i++)
    {
    cin>>a[i];
    if (a[i]>a[i-1] && i!=1)
    {
        t++;
        if (t>max)
        {
            max=t;
        }
    }
    else
     t=0;
    }
     cout<<max+1<<endl;
    return 0; }

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