A - Paper Airplanes CodeForces - 965A (数学)
先算出每个人要用几张纸,然后再算出总共要用几张纸,然后计算用多少包纸
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll k,n,s,p;
cin>>k>>n>>s>>p;
ll t;
if(n%s==0)
t=n/s;
else
t=n/s+1;
ll jk=k*t;
ll ans;
if(jk%p==0)
ans=jk/p;
else
ans=jk/p+1;
cout<<ans;
}
B - Battleship CodeForces - 965B(待填坑)
待填坑
C - Petya and Origami CodeForces - 1080A(数学)
这个肯定都没问题,直接算就行
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll n,k;
cin>>n>>k;
ll red=2*n;
ll green=5*n;
ll blue=8*n;
ll sum=0;
if(red%k==0)
sum+=red/k;
else
sum+=red/k+1;
if(blue%k==0)
sum+=blue/k;
else
sum+=blue/k+1;
if(green%k==0)
sum+=green/k;
else
sum+=green/k+1;
cout<<sum;
}
D - Margarite and the best present CodeForces - 1080B(数学)
数据过大枚举应该是要卡超时
应该去寻找一种直接的数学计算方法
解法
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
ll q;
cin>>q;
while(q--)
{
ll l,r;
cin>>l>>r;
if(l==r)
{
if(l%2!=0)
cout<<-l<<"
";
else
cout<<l<<"
";
}
else
{
if(l%2!=0)
{
ll t=r-l+1;
if(t%2==0)
cout<<(t/2)<<"
";
else
cout<<(t/2)-r<<"
";
}
else
{
ll t=r-l+1;
if(t%2==0)
cout<<-(t/2)<<"
";
else
cout<<r-(t/2)<<"
";
}
}
}
}
E - Masha and two friends CodeForces - 1080C (数学+思维)
这个题实在是太有意思了,太巧妙了。。
解法
代码
#include <bits/stdc++.h>
using namespace std;
long long cdiv(long long x,long long y)
{
return x/y+(x%y>0);
}
long long w(long long a,long long b)
{
return cdiv(a,2)*cdiv(b,2)+(a/2)*(b/2);
}
long long W(long long x1,long long y1,long long x2,long long y2)
{
return w(x2,y2)-w(x2,y1-1)-w(x1-1,y2)+w(x1-1,y1-1);
}
long long B(long long x1,long long y1,long long x2,long long y2)
{
return (y2-y1+1)*(x2-x1+1)-W(x1,y1,x2,y2);
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--)
{
int x1,x2,x3,x4,y1,y2,y3,y4;
long long n,m;
cin>>n>>m;
cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
long long white=W(1,1,m,n);
long long black=B(1,1,m,n);
white+=B(x1,y1,x2,y2);
black-=B(x1,y1,x2,y2);
white-=W(x3,y3,x4,y4);
black+=W(x3,y3,x4,y4);
if(max(x1,x3)<=min(x2,x4)&&max(y1,y3)<=min(y2,y4))
{
white = white - B(max(x1,x3),max(y1,y3),min(x2,x4),min(y2,y4));
black = black + B(max(x1,x3),max(y1,y3),min(x2,x4),min(y2,y4));
}
cout<<white<<" "<<black<<"
";
}
}