Fzu 1076 crossing the desert

Crossing the desert

Time Limit: 1 s	Memory limit: 32 m
Accepted submit205	Total submit304

A jeep arrived at a point on the side of the desert, X kilometers wide. Its fuel consumption was 1 litre/kilometer, and its total assembly volume was 500 litre. Generally, a jeep must use its own fuel tank to set several temporary oil storage points in the desert to cross the desert. If there is enough gasoline available at point A on the edge of the desert, then the jeep will consume at least a few liters of oil from point A through the desert to end B. WriteProgramTo calculate the minimum fuel consumption (accurate to three digits after the decimal point ). (1) assume that the jeep does not fail when walking in the desert; (2) The distance from a point on the desert edge to B is X 500 (that is, the desert width ); Input The first line contains a positive integer k (1 <= k <= 6), indicating the number of test instances. Next to k rows, each row corresponds to a test example, containing a positive integer x, indicating the distance from point A to point B (1 <= x <= 2000 ). Output Each test sample corresponds to a line of output, containing a number, indicating the minimum fuel consumption. Input example 25001000 Output example 500.0003836.497 Original:Fjnu preliminary 2005

Solution:

Reverse Modeling.

500 km before the end point, you can use L of oil directly to the end point. Therefore, the last fuel point is the first 500 kilometers before the end point, and the oil stored here is L. Push forward and backward, to store 500 l in the km, the first fuel point needs to store l, the process round-trip plus the number of one-way past, it is 3. Continue to explore .......

Detailed analysis: http://topic.csdn.net/t/20020906/13/1001713.html

 # Include <iostream> <br/> # include <iomanip> <br/> using namespace STD; </P> <p> int main () <br/>{< br/> int K, I; <br/> double Dist, oil; <br/> while (CIN> K) <br/>{< br/> while (k --) <br/>{< br/> CIN> Dist; <br/> dist-= 500; <br/> oil = 500.0; <br/> I = 1; <br/> while (Dist> 0) <br/>{< br/> dist-= 500.0/(2 * I + 1); <br/> I ++; <br/>}< br/> oil = oil * I + dist * (2 * i-1); <br/> cout <fixed <setprecision (3) <oil <Endl; <br/>}< br/> return 0; <br/>}