POJ基础题（一）

1003 hangover

How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n+ 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

#include<stdio.h>
int sum(double x) {
	int i=2;
	double total=0;
	while (total < x) {
		total += 1.0/ i;
		i++;
	}
	return i-2;
}

int main() {
	int i,t;
	double weight;
	while (scanf("%lf",&weight)&&weight!=0) {
		if (weight > 5.2 || weight < 0)
			return 0;
		else {
			t = sum(weight);
			printf("%d card(s)\n", t);
		}
	}
	return 0;
}

 

 

*1163 The Triangle(动态规划)未A

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right. 

 

#include <stdio.h>
#include <string.h>

int main()
{
	short int a[100][100], i = 0, j = 0, N, max = 0, upleft = 0, upright = 0;
	memset(a, 0, sizeof(a));
	scanf("%d", &N);
	for (i = 0; i < N; i++)
	{
		for (j = 0; j <= i; j++)
		{
			scanf("%d", &a[i][j]);
			short upleft = (i - 1 >= 0) && (j - 1 >= 0) ? a[i - 1][j - 1] : 0;
			short upright = (i - 1 >= 0) ? a[i - 1][j] : 0;
			a[i][j] = (upleft > upright ? upleft : upright) + a[i][j];
			max = a[i][j] > max ? a[i][j] : max;
		}
	}
	printf("%d\n", max);
	return 0;
}

 

1005 I Think I Need a Houseboat

Description

Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land, he learned that the state of Louisiana is actually shrinking by 50 square miles each year, due to erosion caused by the Mississippi River. Since Fred is hoping to live in this house the rest of his life, he needs to know if his land is going to be lost to erosion. 

 

After doing more research, Fred has learned that the land that is being lost forms a semicircle. This semicircle is part of a circle centered at (0,0), with the line that bisects the circle being the X axis. Locations below the X axis are in the water. The semicircle has an area of 0 at the beginning of year 1. (Semicircle illustrated in the Figure.) 

 

Input

The first line of input will be a positive integer indicating how many data sets will be included (N). Each of the next N lines will contain the X and Y Cartesian coordinates of the land Fred is considering. These will be floating point numbers measured in miles. The Y coordinate will be non-negative. (0,0) will not be given.

Output

For each data set, a single line of output should appear. This line should take the form of: “Property N: This property will begin eroding in year Z.” Where N is the data set (counting from 1), and Z is the first year (start from 1) this property will be within the semicircle AT THE END OF YEAR Z. Z must be an integer. After the last data set, this should print out “END OF OUTPUT.”

#include<stdio.h>
#include<math.h>
const static double PI = 3.1415926;
int diffuse(double x, double y) {
	double R2 = x * x + y * y;
	double Area = 0.5 * PI * R2 ;
	return (int)ceil(Area / 50);	// 向上取整
}
int main() {
	int n = 0;	// 测试用例数
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		double x, y;
		scanf("%lf%lf", &x, &y);		
		int cnt = diffuse(x, y);
		printf("Property %d: This property will begin eroding in year %d.\n", i, cnt);
	}
	printf("END OF OUTPUT.\n");
	return 0;
}