Pascal(fpc 3.0.2) 解法, 执行用时: 1ms, 内存消耗: 256K, 提交时间: 2018-08-02 11:55:37

var d1,d2,c,oil,w:extended;
    n,i,j:longint;
    d,p:array[1..100000] of extended;
    f:boolean;
begin
  readln(d1,c,d2,p[0],n);
  for i:=1 to n do
    readln(d[i],p[i]);
  d[n+1]:=d1;
  for i:=0 to n do
  begin
    oil:=oil-(d[i]-d[i-1])/d2;
    if d[i+1]-d[i]>c*d2 then
    begin
      writeln('No Solution');
      halt;
    end;
    f:=false;
    for j:=i+1 to n+1 do
    begin
      if d[j]-d[i]>c*d2 then break;
      if p[j]<p[i] then
      begin
        f:=true;
        break;
      end;
    end;
    if (oil<(d[j]-d[i])/d2) and f then
    begin
      w:=w+((d[j]-d[i])/d2-oil)*p[i];
      oil:=(d[j]-d[i])/d2;
    end
    else if not f then
    begin
      w:=w+(c-oil)*p[i];
      oil:=c;
    end;
  end;
  writeln(w:0:2);
end.

C++(g++ 7.5.0) 解法, 执行用时: 2ms, 内存消耗: 424K, 提交时间: 2023-03-04 17:04:50

#include<bits/stdc++.h>
using namespace std;
double d1,c,d2,p,d[10],v[10],last,ans; 
int n;
int main(){
	scanf("%lf%lf%lf%lf%d",&d1,&c,&d2,&p,&n);
	for(int i=1;i<=n;i++) scanf("%lf%lf",&d[i],&v[i]);
	d[n+1]=d1;v[0]=p;
	for(int i=1;i<=n;i++){		 
		if(d[i+1]-d[i]>c*d2){
			printf("No Solution\n");	
			return 0;
		}
	}
	int j;
	for(int i=0;i<=n;i=j){
		for(j=i+1;j<=n+1;j++){
			if(d[j]-d[i]>c*d2){		
				j--;
				break;
			}
			if(v[j]<v[i]) break;	
		}
		if(v[j]<v[i]){		
			ans+=((d[j]-d[i])/d2-last)*v[i];	 
			last=0;
		}
		else {
			ans+=(c-last)*v[i];			
			last=c-(d[j]-d[i])/d2;
		}
	}
	printf("%.2lf\n",ans);
	return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 3ms, 内存消耗: 480K, 提交时间: 2018-11-09 18:34:39

#include<cstdio>
#define M 100000
int n,i,j;
double o,d1,c,d2,s,m,d[M],p[M];
int main(){
	scanf("%lf%lf%lf%lf%d",&d1,&c,&d2,&p[0],&n);
	d[n+1]=d1;
	for(i=1;i<=n;++i)scanf("%lf%lf",&d[i],&p[i]);
	for(i=1;i<=n+1;++i)
	if(d[i]-d[i-1]>c*d2)return 0*puts("No Solution");
	i=0;
	while(i<=n){
        j=i+1;
        while(d[j]-d[i]<=d2*c&&p[i]<=p[j])++j;
        if(d[j]-d[i]<=d2*c){
            s=d[j]-d[i];
            if(o<s/d2)m+=(s/d2-o)*p[i],o=0;
            else o-=s/d2;
            i=j;
        }
        else{
            m+=(c-o)*p[i];
            o=c-(d[++i]-d[i-1])/d2;
        }
    }
    printf("%0.2lf",m);
	return 0;
}