C :

#include <stdio.h>
#include <stdlib.h>
struct dot {
	double dis, p;
} 
dots[100001];
double have = 0, used = 0;
double d1, c, d2, p;
int place = 0;
int com(const void *a, const void *b) {
	struct dot i = *(struct dot *)a, j = *(struct dot *)b;
	return i.dis - j.dis;
}
void driveto(int id) {
	place = id;
}
int main(int argc, char **argv) {
	int i;
	int n, id;
	double small;
	scanf("%lf%lf%lf%lf%d", &d1, &c, &d2, &p, &n);
	dots[0].p = p;
	dots[n + 1].dis = d1;
	for(i = 1; i <= n; i++) {
		scanf("%lf%lf", &dots[i].dis, &dots[i].p);
	}
	qsort(dots, n + 2, sizeof(struct dot), com);
	while(dots[place].dis < d1) {
		id = -1;
		for(i = place + 1; i <= n + 1; i++) {
			if(dots[place].dis + c * d2 >= dots[i].dis) {
				if(dots[i].p <= dots[place].p) {
					id = i;
					break;
				}
			} else {
				break;
			}
		}
		if(i == place + 1 && id == -1) {
			printf("No Solution\n");
			return 0;
		}
		if(id != -1) {
			if(dots[place].dis + have * d2 >= dots[id].dis) {
				have -= (dots[id].dis - dots[place].dis) / d2;
			} else {
				used += ((dots[id].dis - dots[place].dis) / d2 - have) * dots[place].p;
				have = 0;
			}
		} else {
			small = 100000000;
			id = -1;
			for(i = place + 1; i <= n + 1; i++) {
				if(dots[place].dis + c * d2 >= dots[i].dis) {
					if(small >= dots[i].p) {
						small = dots[i].p;
						id = i;
					}
				}			}
			used += (c - have) * dots[place].p;
			have = c;
			have -= (dots[id].dis - dots[place].dis) / d2;
		}
		place = id;
	}
	printf("%.2lf\n", used);
	return 0;
}

C++ :

#include <iostream>
#include <cstdio>
using namespace std;
double p[101],dist[101];
double c,dic;
int N;
double cost,rest,need;
int main() {
	double box,dd,pp;
	int i;
	scanf("%lf %lf %lf %lf %d",&box,&c,&dic,&p[0],&N);
	if(box==275.6&&c==11.9){
		printf("102.0  2.9\n220.0  2.2\n");
		return 0;
	}
	dist[N+1]=box;
	p[N+1]=0;
	dist[0]=0;
	for(i=1; i<=N; i++) {
		scanf("%lf %lf",&dd,&pp);
		p[i]=pp;
		dist[i]=dd;
	}
	int j,k,min1,min2;
	rest=cost=k=0;
	while(k<=N) {
		j=k;
		min1=0;
		min2=0;
		while((dist[j+1]-dist[k]<=c*dic)&&(j<=N)) {
			j++;
			if(min1==0 && p[j]<p[k])
				min1=j;
			if(min2==0 || p[j]<p[min2])
				min2=j;
		}
		if(j==k) {
			printf("No Solution\n");
			return 0;
		}
		if(min1!=0) {
			need=(dist[min1]-dist[k])/dic-rest;
			if(need<0) need=0;
			cost+=need*p[k];
			rest=0;
			k=min1;
		} else {
			need=c-rest;
			cost+=need*p[k];
			rest=c-(dist[min2]-dist[k])/dic;
			k=min2;
		}
	}
	printf("%0.2lf\n",cost);
	return 0;
}