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直接输出
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pi/ti,for遍历
#include <iostream>
using namespace std;
#define int long long
int a,b,c ;
double t=1.00;
signed main()
{
cin>>a;
int an=0;
for(int i=1;i<=a;i++)
{
cin>>b>>c;
if(t>c*1.00/b)
{
t=c*1.00/b;
an=i;
}
}
cout<<an<<endl;
return 0;
}
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用前缀和求重新启动的次数,特盘端点。
#include <iostream>
using namespace std;
#define int long long
const int n=1e6+11;
int a,b,c,l,r,d[n];
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
cin>>a;
string t;
cin>>t>>b;
t=" "+t;
for(int i=1;i<=a-1;i++)
{
d[i]=d[i-1];
if(t[i]=='1'&&t[i+1]=='0')
{
d[i]++;
}
}
d[a]=d[a-1];
d[a+1]=d[a];
while(b--)
{
cin>>l>>r;
int v=d[r]-d[l-1];
if(r==a+1)
{
cout<<v<<endl;
continue;
}
if(r<=a-1&&t[r]=='1'&&t[r+1]=='0')
{
v--;
}
cout<<v<<endl;
}
}
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背包
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int n=3e3+11;
int a,b[n],c[n],l,r,d[n],f[n];
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
cin>>a;
for(int i=1;i<=a;i++)
{
cin>>b[i];
}
for(int i=1;i<=a;i++)
{
cin>>c[i];
}
for(int i=1;i<=a;i++)
{
cin>>d[i];
}
for(int i=0;i<=3000;i++)
{
f[i]=1e18;
}
f[b[1]]=0;
int v=b[1];
for(int i=1;i<=a;i++)
{
for(int j=b[i]-1;j>=0;j--)
{
f[j]=1e18;
}
for(int j=3000;j>=b[i]&&j>=d[i];j--)
{
f[j]=min(f[j-d[i]]+c[i],f[j]);
}
}
int p=1e18;
for(int i=b[a];i<=3000;i++)
{
p=min(p,f[i]);
}
if(p>=1e18)
{
cout<<"-1"<<endl;
}
else
{
cout<<p<<endl;
}
}
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分别考虑每一位,组合数。
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod=998244353;
const int n=1e6+11;
int a,b[n],c[n],l,r,d[n],f[n];
int ss(int x,int y)
{
int p=1;
while(y)
{
if(y%2==1)
{
p=p*x%mod;
}
x=x*x%mod;
y/=2;
}
return p;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
a=9;
int s=1;
f[1]=1;
for(int i=2;i<=100000;i++)
{
s=s*10%mod;
f[i]=(f[i-1]+s)%mod;
}
s=0;
int ff=0;
for(int i=1;i<=a;i++)
{
cin>>b[i];
if(b[i]!=0)
{
ff++;
}
s+=b[i];
}
if(ff==1)
{
cout<<f[s]%mod<<endl;
return 0;
}
//cout<<"dfd"<<endl;
d[s]=1;
for(int i=s-1;i>=1;i--)
{
d[i]=d[i+1]*i%mod;
}
c[0]=1;
for(int i=1;i<=s-1;i++)
{
c[i]=c[i-1]*i%mod;
}
//cout<<"ff"<<endl;
int an=0;
for(int i=1;i<=a;i++)
{
if(b[i]==0)
continue;
if(b[i]==1)
{
an=(an+(f[s]*i%mod))%mod;
}
else
an=(an+(((d[s-1-(b[i]-1)+1]*ss(c[b[i]-1],mod-2)%mod)*f[s])%mod*i)%mod)%mod;
//cout<<an<<" "<<f[s]<<" "<<d[s-1-(b[i]-1)+1]<<endl;
}
cout<<an<<endl;
}
// 2 3 1 0 0 0 0 0 0