题解

532 条题解

  • 0
    @ 2016-11-18 14:08:36

    这题很简单!queue
    #include<stdio.h>
    #include<stdlib.h>
    #include<queue>
    using namespace std;
    priority_queue<int>que;
    int main(){
    int n;
    scanf("%d",&n);
    for(int i=0,x;i<n;i++){
    scanf("%d",&x);
    que.push(-x);
    }
    int ans=0;
    for(int i=1,tmp;i<n;i++){
    tmp=que.top();
    ans-=que.top();
    que.pop();
    tmp+=que.top();
    ans-=que.top();
    que.pop();
    que.push(tmp);
    }
    printf("%d\n",ans);
    return 0;
    }

  • 0
    @ 2016-11-17 23:11:14

    这是优化过了的C (在电脑无法编译及运行下强行记事本(@——@)
    #include<stdio.h>

    int main()
    {
    int n,i,m,j,t,k,ans=0,a[9999999],x,y;
    scanf("%d",&n);
    m=n;
    for(i=1;i<=n;i++)
    {
    scanf("%d",&a[i]);
    }

    while(m>1)
    {
    if(m==2)
    {
    ans=ans+a[1]+a[2];
    printf("%d",ans);
    return 0;
    }
    x=a[1];
    for(i=2;i<=m;i++)
    {
    if(a[i]<x)
    {
    t=a[i];
    a[i]=x;
    x=t;
    }
    }
    a[1]=x;

    y=a[2];
    for(i=3;i<=m;i++)
    {
    if(a[i]<y)
    {
    t=a[i];
    a[i]=y;
    y=t;
    }
    }
    a[2]=y;

    ans=ans+a[1]+a[2];
    a[1]=a[1]+a[2];
    for(i=2;i<m;i++)
    {
    a[i]=a[i+1];

    }
    m--;

    }

    return 0;
    }

  • 0
    @ 2016-11-16 22:03:43

    为什么用sort函数只得一半而手打快排就得满了?
    #include<iostream>
    #include<algorithm>
    using namespace std;
    bool cmp(int x,int y)
    {
    return x<y;
    }
    int main()
    {
    int n,i,j,count=0,a[10001];
    cin>>n;
    for(i=0;i<n;++i)
    cin>>a[i];
    sort(a,a+n-1,cmp);
    for(i=0;i<n-1;++i)
    {
    count+=a[i]+a[i+1];
    a[i+1]=a[i]+a[i+1];
    a[i]=0;
    for(j=i+1;j<n-1;++j)
    if (a[j]>a[j + 1])
    swap(a[j],a[j + 1]);
    else break;
    }
    cout<<count;
    return 0;
    }

  • 0
    @ 2016-11-14 19:02:07
    #include <bits/stdc++.h>
    using namespace std;
    int n,i,a,b,ans;
    priority_queue<int,vector<int>,greater<int> >q;
    int main() {
        cin>>n;
        for (i=1; i<=n; i++) {
            cin>>a;
            q.push(a);
        }
        while (q.size()>1) {
            a=q.top();
            q.pop();
            b=q.top();
            q.pop();
            ans+=a+b;
            q.push(a+b);
        }
        cout<<ans<<endl;
    }
    
  • 0
    @ 2016-11-09 21:56:14

    注意优先队列默认是大的优先,greater是小的优先,less是大的优先
    #include <cstdio>
    #include <iostream>
    #include <queue>
    using namespace std;
    priority_queue<int,vector<int>,greater<int> > q;

    int main(){
    int n,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
    int t;
    scanf("%d",&t);
    q.push(t);
    }
    for(int i=1;i<=n-1;i++){
    int p1=q.top();
    q.pop();
    int p2=q.top();
    q.pop();
    int t=p1+p2;
    ans+=t;
    q.push(t);
    }
    printf("%d",ans);
    return 0;
    }

    • @ 2016-11-10 15:27:59

      Yes,踩到这个坑了,用自己的数据调了半天发现结果很诡异,然后怒输出中间状态后恍然大悟

  • 0
    @ 2016-11-06 19:31:31

    #include <cstdio>
    #include <cmath>
    #include <algorithm>
    #include <cstring>
    using namespace std;

    long long a[10005];

    int main() {
    int n;
    scanf("%d", &n);
    for (int i=1;i <= n;i++) {
    scanf("%d", &a[i]);
    }
    sort(a+1, a+n+1);

    long long ans=0;
    for (int i=2;i <= n;i++) {
    a[i]+=a[i-1];
    ans+=a[i];
    int k=i;
    while (a[k] > a[k+1] && k+1 <= n) {
    int t=a[k];
    a[k]=a[k+1];
    a[k+1]=t;
    k++;
    }
    }

    printf("%d", ans);

    return 0;
    }

    每次迭代更新,将刚计算出的值更新到递增序列中合适的位置。
    数学归纳法证明,不妨设有三个数(两个数则无需证明,直接合并):a<b<c,则有三种合并方式a+b+a+b+c,a+c+a+b+c,
    b+c+a+b+c,注意到最后三项均为a+b+c,又因为a<b<c,所以三项中最小的是第一个,即先选最小的a,b合并,再与c合并。当又有一项d时可以先选择两项合并,就退化成三项(将合并后的一项看成新的一项)的情况了, 则由数学归纳法就可证明成立。

  • 0
    @ 2016-11-04 10:44:49

    评测结果
    编译成功

    Free Pascal Compiler version 3.0.0 [2015/11/16] for i386
    Copyright (c) 1993-2015 by Florian Klaempfl and others
    Target OS: Win32 for i386
    Compiling foo.pas
    Linking foo.exe
    57 lines compiled, 0.0 sec, 27856 bytes code, 1268 bytes data
    测试数据 #0: Accepted, time = 15 ms, mem = 888 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 888 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 892 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 888 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 888 KiB, score = 10
    测试数据 #5: Accepted, time = 15 ms, mem = 888 KiB, score = 10
    测试数据 #6: Accepted, time = 15 ms, mem = 888 KiB, score = 10
    测试数据 #7: Accepted, time = 0 ms, mem = 888 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 884 KiB, score = 10
    测试数据 #9: Accepted, time = 0 ms, mem = 888 KiB, score = 10
    Accepted, time = 45 ms, mem = 892 KiB, score = 100
    代码
    ```pascal
    var heap:array[0..20001] of longint;
    n,i,t,len,a,ans:longint;

    //function min(a,b:longint):longint;
    //begin if a>b then exit(b);exit(a);end;

    procedure swap(var a,b:longint);
    var t:longint;
    begin t:=a;a:=b;b:=t;end;

    procedure put(a:longint);
    var i:longint;
    begin
    inc(len);
    heap[len]:=a;
    i:=len;
    while (heap[i]<heap[i>>1])and(i<>1) do begin
    swap(heap[i],heap[i>>1]);
    i:=i>>1;
    end;
    end;

    function get:longint;
    var i,t:longint;
    begin
    get:=heap[1];
    heap[1]:=heap[len];
    heap[len]:=maxlongint;
    dec(len);i:=1;
    if len=0 then exit;
    while true do begin
    t:=i<<1;
    if (heap[i]>heap[t])and(heap[t]<heap[t+1])then begin
    swap(heap[i],heap[t]);i:=t;end
    else if (heap[i]>heap[t+1])then begin
    swap(heap[i],heap[t+1]);i:=t+1;end
    else break;
    end;
    end;

    begin
    readln(n);
    filldword(heap,length(heap),maxlongint);
    //writeln(heap[1]);
    len:=0;
    for i:=1 to n do begin
    read(t);
    put(t);
    end;
    ans:=0;//for i:=1 to len do write(heap[i],' ');writeln;
    while len>1 do begin
    a:=get+get;
    inc(ans,a);
    put(a);
    //for i:=1 to len do write(heap[i],' ');writeln;
    end;
    writeln(ans);
    end.
    ```

  • 0
    @ 2016-10-28 20:31:19
    var
      n,i,s:longint;
      a:array[0..11111] of longint;
    procedure qsort(l,r:longint);
      var
        i,j,x,p:longint;
      begin
        i:=l;
        j:=r;
        x:=random(j-i+1)+i;
        p:=a[x];
        repeat
          while a[i]>p do inc(i);
          while a[j]<p do dec(j);
          if i<=j
            then
              begin
                a[0]:=a[i];
                a[i]:=a[j];
                a[j]:=a[0];
                inc(i);
                dec(j);
              end;
        until i>j;
        if i<r then qsort(i,r);
        if j>l then qsort(l,j);
      end;
    procedure insort(i:longint);
      var
        h:longint;
      begin
        a[0]:=a[i];
        h:=i-1;
        while a[0]>a[h] do
          begin
            a[h+1]:=a[h];
            dec(h);
          end;
        a[h+1]:=a[0];
      end;
    begin
      randomize;
      readln(n);
      for i:=1 to n do read(a[i]);
      readln;
      qsort(1,n);
      for i:=n-1 downto 1 do
        begin
          a[i]:=a[i+1]+a[i];
          s:=s+a[i];
          insort(i);
        end;
      writeln(s);
      readln;
    end.
    
  • 0
    @ 2016-10-20 19:24:57
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int N=30005;
    int n,a[N],ans(0);
    void adjust(int i,int k){
        int j=2*i+1;
        while(j<=k){ 
            if(j<=k){
                if(j<k&&a[j]>a[j+1])j++; 
                if(a[i]>a[j]){
                    swap(a[i],a[j]);
                    i=j;
                    j=2*i+1;
                }
                else break; 
            }
        }
    }
    void heapsort(int n){/*将二叉树调整成小顶堆*/
        for(int i=(n-2)/2;i>=0;i--)adjust(i,n-1);
        for(int i=n-1;i>0;i--){
            swap(a[0],a[i]);
            adjust(0,i-1);
            a[0]+=a[i];ans+=a[0];
            adjust(0,i-1);
        }
    }
    int main(){
        cin>>n;
        for(int i=0;i<n;i++)cin>>a[i];
        heapsort(n);
        cout<<ans<<endl;
        return 0; 
    }
    
  • 0
    @ 2016-10-20 19:24:05
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    int f[60007];
    
    int main()
    {
        int n, i;
        cin >> n;
        for (i = 1; i <= n; i++) cin >> f[i];
        
        int m = (n << 1) - 1, Res = 0, t1 = 1, t2 = n + 1, t, s[2] = {0};
        sort(f + 1, f + n + 1);
        for (i = n + 1; i <= m; i++)
        {
            for (t = 0; t < 2; t++)
                if ((t2 < i && f[t2] < f[t1]) || t1 > n) s[t] = f[t2++];
                else s[t] = f[t1++];
            Res += (f[i] = s[0] + s[1]);
        }
        
        cout << Res << endl;
        
        return 0;
    }
    
    
  • 0
    @ 2016-10-20 19:22:20
    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int N=30005;
    int n,a[N],ans(0);
    int main(){
        cin>>n;
        for(int i=0;i<n;i++)cin>>a[i];
        make_heap(a,a+n,greater<int>()); 
        for(int i=n-1;i>0;i--){
            pop_heap(a,a+i+1,greater<int>());
            pop_heap(a,a+i,greater<int>());
            a[i-1]+=a[i];
            ans+=a[i-1];
            push_heap(a,a+i,greater<int>());
        }
        cout<<ans<<endl;
        return 0; 
    }
    
  • 0
    @ 2016-10-07 16:13:36

    //fruit
    //2016-10-06 Shanghaineese Eccentric
    #include<iostream>
    #include<cstdio>
    #include<queue>
    using namespace std;
    priority_queue<int> que;
    int n,sum;

    int main()
    {
    int a,b,i,temp;
    // freopen("fruit2.in","r",stdin);
    // freopen("fruit.out","w",stdout);
    sum=0;
    scanf("%d",&n);
    for (i=0;i<n;i++)
    {
    scanf("%d",&temp);
    que.push(-temp);
    }
    while (que.size()>=2)
    {
    a=que.top();
    que.pop();
    b=que.top();
    que.pop();
    sum+=a+b;
    que.push(a+b);
    }
    cout<<-sum<<endl;
    return(0);
    }

  • 0
    @ 2016-10-02 16:19:44

    #include <iostream>

    using namespace std;

    int apple[10000];

    void ksort(int l, int r)//快排
    {
    if (l == r)return;
    int mid = apple[(l + r) / 2];
    int i = l, j = r;
    int temp;
    do
    {
    while (apple[i] < mid)i++;
    while (mid < apple[j])j--;
    if (i <= j)
    {
    temp = apple[i];
    apple[i] = apple[j];
    apple[j] = temp;
    i++;
    j--;
    }
    } while (i <= j);
    if (i <= r)ksort(i, r);
    if (l <= j)ksort(l, j);
    return;
    }

    int main()
    {
    int ans = 0;//花费力气
    int n;//堆数
    int i, j;
    cin >> n;
    for (i = 0; i < n; ++i)
    cin >> apple[i];
    ksort(0, n - 1);
    for (i = 0; i < n-1; ++i)
    {
    ans += apple[i] + apple[i + 1];
    apple[i + 1] = apple[i] + apple[i + 1];
    apple[i] = 0;
    for (j = i + 1; j < n - 1; ++j)
    if (apple[j] > apple[j + 1])swap(apple[j], apple[j + 1]);
    else break;
    }
    cout << ans;
    system("pause");
    return 0;
    }

  • 0
    @ 2016-10-02 16:19:39

    #include <iostream>

    using namespace std;

    int apple[10000];

    void ksort(int l, int r)//快排
    {
    if (l == r)return;
    int mid = apple[(l + r) / 2];
    int i = l, j = r;
    int temp;
    do
    {
    while (apple[i] < mid)i++;
    while (mid < apple[j])j--;
    if (i <= j)
    {
    temp = apple[i];
    apple[i] = apple[j];
    apple[j] = temp;
    i++;
    j--;
    }
    } while (i <= j);
    if (i <= r)ksort(i, r);
    if (l <= j)ksort(l, j);
    return;
    }

    int main()
    {
    int ans = 0;//花费力气
    int n;//堆数
    int i, j;
    cin >> n;
    for (i = 0; i < n; ++i)
    cin >> apple[i];
    ksort(0, n - 1);
    for (i = 0; i < n-1; ++i)
    {
    ans += apple[i] + apple[i + 1];
    apple[i + 1] = apple[i] + apple[i + 1];
    apple[i] = 0;
    for (j = i + 1; j < n - 1; ++j)
    if (apple[j] > apple[j + 1])swap(apple[j], apple[j + 1]);
    else break;
    }
    cout << ans;
    system("pause");
    return 0;
    }

  • 0
    @ 2016-09-23 23:08:24
    #include<bits/stdc++.h>
    using namespace std;
    struct cmp1    
    {      
        bool operator ()(int &a,int &b)    
        {      
            return a>b;
        }      
    }; 
    priority_queue<int,vector<int>,cmp1>x;
    int n,a,num=0,b;
    int main(){
    freopen("hb.in","r",stdin);
    freopen("hb.out","w",stdout);
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a);
        x.push(a);
    }
    for( int j=0;j<n-1;j++){
        a=x.top();
        x.pop();
        b=x.top();
        x.pop();
        x.push(a+b);
        num+=a+b;
        
    }
    cout<<num;
    return 0;
    }
    
  • 0
    @ 2016-08-31 21:32:44

    STL大法

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<queue>
    using namespace std;
    int a;
    int n,l;
    long long f1,f2,ans=0;
    int input()
    {
    int x = 0;
    int f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
    if (ch == '-')
    f = -1;
    ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
    x = x * 10 + ch - '0';
    ch = getchar();
    }
    return x * f;
    }
    struct cmp{
    bool operator() (const int a,const int b) const{
    return a > b;
    }
    };
    priority_queue<int,vector<int>,cmp> q;
    int main()
    {
    n=input();
    for(int i=1;i<=n;i++)
    {
    a=input();
    q.push(a);
    }
    while(q.size()!=1)
    {
    f1=q.top();
    q.pop();
    f2=q.top();
    q.pop();
    ans+=f1+f2;
    q.push(f1+f2);
    }
    cout<<ans;
    return 0;
    }

  • 0
    @ 2016-08-27 07:22:50

    #include"iostream"
    #include"stdio.h"
    #include"vector"
    #include"functional"
    #include"string"
    #include"cstring"
    #include"algorithm"
    #include"queue"
    #include"set"
    #include"queue"
    #define f1 cout<<"fuck1"<<endl;
    #define f2 cout<<"fuck2"<<endl;

    #define f(x) (p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u)
    #define eps=1e-6
    using namespace std;
    typedef pair<int,int> pii;
    typedef vector<int> vi;
    int main()
    {
    int n;
    while(cin>>n)
    {
    priority_queue<int,vector<int>,greater<int> >Q;
    int i;
    for(i=0;i<n;i++)
    {
    int a;
    scanf("%d",&a);
    Q.push(a);
    }
    int ans=0;
    while(Q.size()!=1)
    {
    int t1,t2;
    t1=Q.top();Q.pop();t2=Q.top();Q.pop();
    ans+=t1+t2;
    Q.push(t1+t2);
    }
    cout<<ans<<endl;
    }
    }

  • 0
    @ 2016-08-24 20:42:02

    用c++stl库就行啦

    #include<cstdio>
    #include<iostream>
    #include<queue>
    using namespace std;
    long long n,x,y,ans=0;
    priority_queue<long long,vector<long long>,greater<long long> > h;
    int main()
    {
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
    {
    scanf("%lld",&x);
    h.push(x);
    }
    for(int i=1;i<=n-1;i++)
    {
    x=h.top();h.pop();
    y=h.top();h.pop();
    ans+=x+y;h.push(x+y);
    }
    printf("%lld",ans);
    return 0;
    }

  • 0
    @ 2016-08-16 10:16:10

    program fruit;
    var
    a:array [1..10000] of longint;
    i,m,n,x,j,total:longint;

    procedure QuickSort(low,high:longint);
    var
    i,j,mid,med:longint;
    begin
    i:=low;
    j:=high;
    mid:=a[(i+j) div 2];
    repeat
    while a[i]<mid do
    i:=i+1;
    while a[j]>mid do
    j:=j-1;
    if i<=j then
    begin
    med:=a[i];
    a[i]:=a[j];
    a[j]:=med;
    i:=i+1;
    j:=j-1;
    end;
    until i>j;
    if i<high then QuickSort(i,high);
    if j>low then QuickSort(low,j);
    end;

    begin
    readln(n);
    for i:=1 to n do
    read(a[i]);
    QuickSort(1,n);
    for i:=1 to n-1 do
    begin
    x:=a[i]+a[i+1];
    a[i]:=0;
    a[i+1]:=0;
    total:=total+x;
    j:=i+2;
    while (j<=n) and (x>a[j]) do
    j:=j+1;
    j:=j-1;
    for m:=i+1 to j do
    a[m]:=a[m+1];
    a[j]:=x;
    end;
    write(total);
    end.

  • 0
    @ 2016-08-15 00:30:02
    #include <bits/stdc++.h>
    #define maxN 100010
    
    using namespace std;
    typedef long long QAQ ;
    
    int arr[maxN];
    
    priority_queue<int ,vector<int>,greater<int> >Q; 
    QAQ ans ;
    int main(){
        int n,tmp2,tmp1;
        ios::sync_with_stdio(false);
        cin >> n ;
        for(int i=1 ; i<=n ;++i)cin >> arr[i];
        sort(arr+1,arr+n+1);
        for(int i=1;i<=n;++i){
            Q.push(arr[i]);
        }
        for(int i=1;i<=n-1;++i){
            tmp1=Q.top();
            Q.pop();
            tmp2=Q.top();
            Q.pop();
            Q.push(tmp1+tmp2);
            ans+=tmp1+tmp2;
        }
        printf("%lld",ans);
        return 0 ;
    } 
    

信息

ID
1097
难度
6
分类
贪心 点击显示
标签
递交数
23854
已通过
6310
通过率
26%
被复制
41
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