标签：style os io for ar 代码 时间 line
这道题是树状数组的题，但是用普通数组也能整出来，没学树状数组，就用的普通数组，算是高效算法吧，下面是我的修改思路：
1.一上来我写了如下代码：把每个新加进去的数压入vector，但是我在当t=1时，我的想法是将v[i]一个一个加上去，这样肯定会超时，而且我这个方法的sum是最后一起求的，这样多了个循环，又耗费时间。
#include<stdio.h>
#include<time.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
vector<int> v;
v.push_back(0);
int n;cin>>n;
while(n--)
{
int t;cin>>t;
if(t == 1)
{
int a,x;cin>>a>>x;
for(int i = 0;i < a;i++)
v[i] += x;
}
if(t == 2)
{
int k;cin>>k;
v.push_back(k);
}
if(t == 3)
v.pop_back();
int sum = 0;
for(int i = 0;i < v.size();i++)
sum += v[i];
printf("%.6lf\n",(double)sum*1.0/v.size());
}
return 0;
} 2.想了一下，做出以下修改：虽然我意识到了sum用成全局的，随输随加，但其余的真是纯属脑残，因为我没考虑到连续把最后一个数删除的情况，看来输进去的数还是要用vector保存
#include<stdio.h>
#include<time.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
int n,last = 0,shu = 1,a,k,x,sum = 0;cin>>n;
while(n--)
{
int t;cin>>t;
if(t == 1) {cin>>a>>x;sum += a*x;}
if(t == 2) {cin>>k;sum += k;last = k;shu++;}
if(t == 3) {sum -= k;shu--;}
printf("%.6lf\n",sum*1.0/shu);
}
return 0;
}#include<stdio.h>
#include<time.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
vector<int> v;
v.push_back(0);
int n,last = 0,shu = 1,a,k,x,sum = 0;cin>>n;
while(n--)
{
int t;cin>>t;
if(t == 1)
{
cin>>a>>x;sum += a*x;
for(int i = 0;i < a;i++)
v[i] += x;
}
if(t == 2)
{cin>>k;sum += k;v.push_back(k);}
if(t == 3)
{sum -= v[v.size() - 1];v.pop_back();}
printf("%.6lf\n",(double)sum*1.0/v.size());
}
return 0;
}#include<stdio.h>
#include<time.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
vector<int> v;
v.push_back(0);
int n,last = 0,shu = 1,k,a,x,sum = 0,biao[200005];cin>>n;
memset(biao,0,sizeof(biao));
while(n--)
{
int t;cin>>t;
if(t == 1)
{
cin>>a>>x;
for(int i = 0;i < a;i++)
biao[i] += x;
sum += a*x;
}
if(t == 2)
{cin>>k;sum += k;v.push_back(k);}
if(t == 3)
{
sum -= v[v.size() - 1];
sum -= biao[v.size() - 1];
biao[v.size() - 1] = 0;
v.pop_back();
}
printf("%.6lf\n",(double)sum*1.0/v.size());
}
return 0;
}5.经过参考网上的题解，我发现这题竟然用long long！就修改成long long
#include<stdio.h>
#include<time.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
vector<int> v;
v.push_back(0);
int n,last = 0,shu = 1,k,a,x,biao[200005];cin>>n;
long long sum = 0;
memset(biao,0,sizeof(biao));
while(n--)
{
int t;cin>>t;
if(t == 1)
{
cin>>a>>x;
for(int i = 0;i < a;i++)
biao[i] += x;
sum += a*x;
}
if(t == 2)
{cin>>k;sum += k;v.push_back(k);}
if(t == 3)
{
sum -= v[v.size() - 1];
sum -= biao[v.size() - 1];
biao[v.size() - 1] = 0;
v.pop_back();
}
printf("%.6lf\n",(double)sum*1.0/v.size());
}
return 0;
}#include<stdio.h> #include<time.h> #include<string.h> #include<string> #include<stdlib.h> #include<iostream> #include<vector> using namespace std; int main() { vector<int> v; v.push_back(0); int n,last = 0,shu = 1,k,a,x,biao[200005];cin>>n; long long sum = 0; while(n--) { int t;cin>>t; if(t == 1) { cin>>a>>x; biao[a - 1] += x; sum += a*x; } if(t == 2) {cin>>k;sum += k;v.push_back(k);} if(t == 3) { sum -= v[v.size() - 1]; sum -= biao[v.size() - 1]; biao[v.size() - 2] += biao[v.size() - 1]; biao[v.size() - 1] = 0; v.pop_back(); } printf("%.6lf\n",(double)sum*1.0/v.size()); } return 0; }
Codeforces-Round 174(Cows and Sequence),布布扣,bubuko.com
Codeforces-Round 174(Cows and Sequence)
标签：style os io for ar 代码 时间 line
原文地址：http://blog.csdn.net/glqglqglq2/article/details/38436957
