Codeforces-Round 174(Cows and Sequence)(树状数组、高校算法)

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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;}

3.又修改：这样去除了sum的循环，但是当t=1时循环还在，还TLE

#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;}

4.无奈之下开始打表，把每个数的位置对应的加的数都打到表里，无奈有挂了，还是WA

#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;}

6.修改后又挂了，TLE，发现这里有点小技巧，t=1时根本不用将biao数组的所有前a个值都加x，将a加上x即可，然后从t=3里面调节，太巧妙了，终于AC

#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;}

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