A题

题意:将一个非递增数组的某个元素往前移至少k个单位,然后求哪种情况下移动之后数组与位权乘积之和最小。
思路:前缀和+贪心+枚举。暴力解法肯定被T,可以证明,答案一定要在只能往前移恰好k步的时候取得。这个可以通过简单地模拟一下,因为这是非递增地序列,每往前移动一步,比如将a[i]移到a[i-1],那么就相当于是总的和-a[i],再+a[i-1],所以可得你每移动一步,你的和一定是减小或不变的,那就要恰好移动最少的步数k步就行了。往前移k步,假如在没有移动之前它的总和为sum,那么你将某个数x往前移动k个单位的话,那么你的sum就要在原来地基础上-xk,即sum-kx,然后,在你将这个数插入的位置处的后面k个数他们的位权都是+1了,假设这后面k个数字之和为y,则最后的sum就是sum=sum-k*x+y.推出了这个式子之后,我们就要最数组做一个预处理,即从第k+1个数开始,记录下它前面k个数字的和,并用数组存起来。

for(int i=1;i<=n;i++){
        sum_k+=a[i];
        if(i>k){
            sum_k-=a[i];
            more[i]=sum_k;
            sum_k+=a[i];
            sum_k-=a[i-k];
        }
}

就这样,按照这种写法,你遍历到了哪一个,你sum_k先减去这个数字,然后存在more数组里即可。
最后,你只要一次遍历这个数组,然后记录下减少量最小的值就行了。

#include<iostream>
using namespace std;
const int MAX=100010;
typedef long long ll;
ll a[MAX],more[MAX];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n,k;
        scanf("%ld%ld",&n,&k);
        ll sum=0;
        for(ll i=1;i<=n;i++){
            scanf("%ld",&a[i]);
            sum+=(i*a[i]);
        }
        ll sum_k=0;
        for(int i=1;i<=n;i++){  //预处理,记录下每个数的前k个数之和
            sum_k+=a[i];
            if(i>k){
                sum_k-=a[i];
                more[i]=sum_k;
                sum_k+=a[i];
                sum_k-=a[i-k];
            }
        }
        ll ans=1e18;
        for(ll i=k+1;i<=n;i++){
            ll less=k*a[i];  //少了多少 
            ll temp=less-more[i];  //少的-多的
            if(ans>temp)  ans=temp;
        }
        ans=sum-ans;
        printf("%ld\n",ans);
    }
    return 0;
}

B题

题意:就是有一列人在跑步,然后按随机的序列,求每个人都到达队首的所用时间的期望。
思路:枚举,这个题目看似复杂,实则是纸老虎,因为是随机的,所以可以求出每个人在任意一个位置跑到队首的时间总和除以n即可。而且,可以发现,每个配送员的跑的路程都是一样的。

#include<bits/stdc++.h>
using namespace std;
double c[1010],d[1010];
int main(){
    int n;
    double v,u;
    cin>>n>>v>>u;
    for(int i=1;i<=n;++i)  cin>>c[i];
    for(int i=1;i<=n;++i)   cin>>d[i];
    double ans=0;
    double sum=u*n;
    for(int i=1;i<=n;++i){  //第i个配送员
        for(int j=1;j<=n;++j){  //以第几的顺序跑到队首
            double t=sum/(c[i]-(j-1)*d[i]-v);  //记下速度的差值,然后用总路程除它即可求出在该情况的时间
            ans+=t;
        }
    }
    ans/=n;
    cout<<fixed<<setprecision(3)<<ans<<endl;
    return 0;
}

E题

题意:略
思路:打表+枚举即可。具体的是先写一个程序来将在最大范围内的所有符合的数都求出来,然后粘贴下来放到一个数组里面。(这里最好也记录一下有多个满足的数字),而且,还要注意,最后一个数字是4444444444(10个4),要超过1e9。存到一个数组后,那么就从左边遍历就可以了,不过这里还可以优化一下
看这里:

while(1){
        ll ind=lower_bound(a,a+1023,l)-a;
        ll val=a[ind];
        if(val>=r){
            ans+=(r-l+1)*val;
            break;
        }
        else{
            ans+=(val-l+1)*val;
            l=val+1;
        }
}

我们每次利用lower_bound函数将大于左边的第一个满足的数字找出,为val,然后val如果比r更大的话,那么说明区间长度为r-l+1,否则区间长度就是val-l+1,在乘以相应的权即可。(密集恐惧症人群请回避

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[1023]={4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,
4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,
44444,44447,44474,44477,44744,44747,44774,44777,47444,47447,47474,47477,
47744,47747,47774,47777,74444,74447,74474,74477,74744,74747,74774,74777,
77444,77447,77474,77477,77744,77747,77774,77777,444444,444447,444474,444477,
444744,444747,444774,444777,447444,447447,447474,447477,447744,447747,447774,
447777,474444,474447,474474,474477,474744,474747,474774,474777,477444,477447,
477474,477477,477744,477747,477774,477777,744444,744447,744474,744477,744744,
744747,744774,744777,747444,747447,747474,747477,747744,747747,747774,747777,
774444,774447,774474,774477,774744,774747,774774,774777,777444,777447,777474,
777477,777744,777747,777774,777777,4444444,4444447,4444474,4444477,4444744,4444747,
4444774,4444777,4447444,4447447,4447474,4447477,4447744,4447747,4447774,4447777,4474444,
4474447,4474474,4474477,4474744,4474747,4474774,4474777,4477444,4477447,4477474,4477477,
4477744,4477747,4477774,4477777,4744444,4744447,4744474,4744477,4744744,4744747,4744774,
4744777,4747444,4747447,4747474,4747477,4747744,4747747,4747774,4747777,4774444,4774447,
4774474,4774477,4774744,4774747,4774774,4774777,4777444,4777447,4777474,4777477,4777744,
4777747,4777774,4777777,7444444,7444447,7444474,7444477,7444744,7444747,7444774,7444777,
7447444,7447447,7447474,7447477,7447744,7447747,7447774,7447777,7474444,7474447,7474474,
7474477,7474744,7474747,7474774,7474777,7477444,7477447,7477474,7477477,7477744,7477747,
7477774,7477777,7744444,7744447,7744474,7744477,7744744,7744747,7744774,7744777,7747444,
7747447,7747474,7747477,7747744,7747747,7747774,7747777,7774444,7774447,7774474,7774477,
7774744,7774747,7774774,7774777,7777444,7777447,7777474,7777477,7777744,7777747,7777774,
7777777,44444444,44444447,44444474,44444477,44444744,44444747,44444774,44444777,44447444,
44447447,44447474,44447477,44447744,44447747,44447774,44447777,44474444,44474447,44474474,
44474477,44474744,44474747,44474774,44474777,44477444,44477447,44477474,44477477,44477744,
44477747,44477774,44477777,44744444,44744447,44744474,44744477,44744744,44744747,44744774,
44744777,44747444,44747447,44747474,44747477,44747744,44747747,44747774,44747777,44774444,
44774447,44774474,44774477,44774744,44774747,44774774,44774777,44777444,44777447,44777474,
44777477,44777744,44777747,44777774,44777777,47444444,47444447,47444474,47444477,47444744,
47444747,47444774,47444777,47447444,47447447,47447474,47447477,47447744,47447747,47447774,
47447777,47474444,47474447,47474474,47474477,47474744,47474747,47474774,47474777,47477444,
47477447,47477474,47477477,47477744,47477747,47477774,47477777,47744444,47744447,47744474,
47744477,47744744,47744747,47744774,47744777,47747444,47747447,47747474,47747477,47747744,
47747747,47747774,47747777,47774444,47774447,47774474,47774477,47774744,47774747,47774774,
47774777,47777444,47777447,47777474,47777477,47777744,47777747,47777774,47777777,74444444,
74444447,74444474,74444477,74444744,74444747,74444774,74444777,74447444,74447447,74447474,
74447477,74447744,74447747,74447774,74447777,74474444,74474447,74474474,74474477,74474744,
74474747,74474774,74474777,74477444,74477447,74477474,74477477,74477744,74477747,74477774,
74477777,74744444,74744447,74744474,74744477,74744744,74744747,74744774,74744777,74747444,
74747447,74747474,74747477,74747744,74747747,74747774,74747777,74774444,74774447,74774474,
74774477,74774744,74774747,74774774,74774777,74777444,74777447,74777474,74777477,74777744,
74777747,74777774,74777777,77444444,77444447,77444474,77444477,77444744,77444747,77444774,
77444777,77447444,77447447,77447474,77447477,77447744,77447747,77447774,77447777,77474444,
77474447,77474474,77474477,77474744,77474747,77474774,77474777,77477444,77477447,77477474,
77477477,77477744,77477747,77477774,77477777,77744444,77744447,77744474,77744477,77744744,
77744747,77744774,77744777,77747444,77747447,77747474,77747477,77747744,77747747,77747774,
77747777,77774444,77774447,77774474,77774477,77774744,77774747,77774774,77774777,77777444,
77777447,77777474,77777477,77777744,77777747,77777774,77777777,444444444,444444447,444444474,
444444477,444444744,444444747,444444774,444444777,444447444,444447447,444447474,444447477,
444447744,444447747,444447774,444447777,444474444,444474447,444474474,444474477,444474744,
444474747,444474774,444474777,444477444,444477447,444477474,444477477,444477744,444477747,
444477774,444477777,444744444,444744447,444744474,444744477,444744744,444744747,444744774,
444744777,444747444,444747447,444747474,444747477,444747744,444747747,444747774,444747777,
444774444,444774447,444774474,444774477,444774744,444774747,444774774,444774777,444777444,
444777447,444777474,444777477,444777744,444777747,444777774,444777777,447444444,447444447,
447444474,447444477,447444744,447444747,447444774,447444777,447447444,447447447,447447474,
447447477,447447744,447447747,447447774,447447777,447474444,447474447,447474474,447474477,
447474744,447474747,447474774,447474777,447477444,447477447,447477474,447477477,447477744,
447477747,447477774,447477777,447744444,447744447,447744474,447744477,447744744,447744747,
447744774,447744777,447747444,447747447,447747474,447747477,447747744,447747747,447747774,
447747777,447774444,447774447,447774474,447774477,447774744,447774747,447774774,447774777,
447777444,447777447,447777474,447777477,447777744,447777747,447777774,447777777,474444444,
474444447,474444474,474444477,474444744,474444747,474444774,474444777,474447444,474447447,
474447474,474447477,474447744,474447747,474447774,474447777,474474444,474474447,474474474,
474474477,474474744,474474747,474474774,474474777,474477444,474477447,474477474,474477477,
474477744,474477747,474477774,474477777,474744444,474744447,474744474,474744477,474744744,
474744747,474744774,474744777,474747444,474747447,474747474,474747477,474747744,474747747,
474747774,474747777,474774444,474774447,474774474,474774477,474774744,474774747,474774774,
474774777,474777444,474777447,474777474,474777477,474777744,474777747,474777774,474777777,
477444444,477444447,477444474,477444477,477444744,477444747,477444774,477444777,477447444,
477447447,477447474,477447477,477447744,477447747,477447774,477447777,477474444,477474447,
477474474,477474477,477474744,477474747,477474774,477474777,477477444,477477447,477477474,
477477477,477477744,477477747,477477774,477477777,477744444,477744447,477744474,477744477,
477744744,477744747,477744774,477744777,477747444,477747447,477747474,477747477,477747744,477747747,477747774,477747777,477774444,477774447,477774474,477774477,477774744,477774747,477774774,477774777,477777444,477777447,477777474,477777477,477777744,477777747,477777774,477777777,744444444,744444447,744444474,744444477,744444744,744444747,744444774,744444777,744447444,744447447,744447474,744447477,744447744,744447747,744447774,744447777,744474444,744474447,744474474,744474477,744474744,744474747,744474774,744474777,744477444,744477447,744477474,744477477,744477744,744477747,744477774,744477777,744744444,744744447,744744474,744744477,744744744,744744747,744744774,744744777,744747444,744747447,744747474,744747477,744747744,744747747,744747774,744747777,744774444,744774447,744774474,744774477,744774744,744774747,744774774,744774777,744777444,744777447,744777474,744777477,744777744,744777747,744777774,744777777,747444444,747444447,747444474,747444477,747444744,747444747,747444774,747444777,747447444,747447447,747447474,747447477,747447744,747447747,747447774,747447777,747474444,747474447,747474474,747474477,747474744,747474747,747474774,747474777,747477444,747477447,747477474,747477477,747477744,747477747,747477774,747477777,747744444,747744447,747744474,747744477,747744744,747744747,747744774,747744777,747747444,747747447,747747474,747747477,747747744,747747747,747747774,747747777,747774444,747774447,747774474,747774477,747774744,747774747,747774774,747774777,747777444,747777447,747777474,747777477,747777744,747777747,747777774,747777777,774444444,774444447,774444474,774444477,774444744,774444747,774444774,774444777,774447444,774447447,774447474,774447477,774447744,774447747,774447774,774447777,774474444,774474447,774474474,774474477,774474744,774474747,774474774,774474777,774477444,774477447,774477474,774477477,774477744,774477747,774477774,774477777,774744444,774744447,774744474,774744477,774744744,774744747,774744774,774744777,774747444,774747447,774747474,774747477,774747744,774747747,774747774,774747777,774774444,774774447,774774474,774774477,774774744,774774747,774774774,774774777,774777444,774777447,774777474,774777477,774777744,774777747,774777774,774777777,777444444,777444447,777444474,777444477,777444744,777444747,777444774,777444777,777447444,777447447,777447474,777447477,777447744,777447747,777447774,777447777,777474444,777474447,777474474,777474477,777474744,777474747,777474774,777474777,777477444,777477447,777477474,777477477,777477744,777477747,777477774,777477777,777744444,777744447,777744474,777744477,777744744,777744747,777744774,777744777,777747444,777747447,777747474,777747477,777747744,777747747,777747774,777747777,777774444,777774447,777774474,777774477,777774744,777774747,777774774,777774777,777777444,777777447,777777474,777777477,777777744,777777747,777777774,777777777,4444444444};
int k=0;
int main(){
    ll l,r;
    cin>>l>>r;
    ll ans=0;
    while(1){
        ll ind=lower_bound(a,a+1023,l)-a;  //找到第一个大于左区间的值
        ll val=a[ind];
        if(val>=r){  //如果val超过r
            ans+=(r-l+1)*val;  //区间长为r-l+1
            break;
        }
        else{
            ans+=(val-l+1)*val;  //否则为val-l+1
            l=val+1;
        }
    }
cout<<ans<<endl;
    return 0;
}