With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked…

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked…

找出最小开销. 思路: 出发点的加油站编号设为0,终点的加油站编号设为n,其他加油站编号按距离依次排序. 如果0号加油站的距离!=0,则无法出发,行驶距离为0. 从起点开始,寻找规则为,如果存在油价小于本加油站的油价的,则计入, 没有就计入油价最低的. 如此循环,如果能到达终点,输出总花销:不能,输出总行驶距离. ps:输出的字符的拼写不能有误. #include<cstdio> #include<iostream> #include<algorithm> using…

一.技术总结 是贪心算法的题目,题目主要考虑的问题有几个,是否会在第一个加油站的最近距离大于0,如果是这样那么直接输出答案,因为初始油箱没有汽油: 第二个是如何选定加油站,如果在可到达距离范围类,我们优先考虑比当前加油站价格更低的,然后如果有,就直接到达这里,如果没有那也要选出这里面价格最低的那个加油站, 然后在当前加油站,加满油箱.这样可以更加的省钱,那么油箱会多出油行驶距离(leftdis = Cmax*Davg -(minPriceDis - nowdis)). 可以开始判断最后结局了,如…

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked…

1033 To Fill or Not to Fill (25 分) With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may g…

Source: PAT A1033 To Fill or Not to Fill (25 分) Description: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. D…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…