UVa 485 - Pascal's Triangle of Death

題目：打印Pascal三角到第一个到达10^60的行。

分析：字符串、大整數、模擬。f（i，j）= f（i-1，j-1）+ f（i-1，j） {組合數公式}。

說明：注意不小于10^60的數字有61位(⊙_⊙)。

#include <iostream>
#include <cstdlib>
#include <cstdio>

using namespace std;

int P[220][220][70] = {0};

int main()
{
	for (int i = 0 ; i < 210 ; ++ i)
		P[i][0][0] = P[i][i][0] = 1;
	printf("1\n");
	for (int i = 1 ; i < 210 ; ++ i) {
		int end = 0,max = 0;
		printf("1 ");
		for (int j = 1 ; j < i  ; ++ j) {
			for (int k = 0 ; k < 65 ; ++ k) {
				P[i][j][k] += P[i-1][j][k]+P[i-1][j-1][k];
				if ( P[i][j][k] > 9 ) {
					P[i][j][k+1] += P[i][j][k]/10;
					P[i][j][k] %= 10;
				}
			}
			end = 64;
			while (!P[i][j][end]) -- end;
			if (max < end) max = end;
			while (end >= 0) printf("%d",P[i][j][end --]);
			printf(" ");
		}
		printf("1\n");
		if (max >= 60) break;
	}
	return 0;
}

时间： 2024-12-15 17:14:26

UVa 485 - Pascal's Triangle of Death的相关文章

（leetcode题解）Pascal's Triangle

Pascal's Triangle Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 题意实现一个杨辉三角. 这道题只要注意了边界条件应该很好实现出来,C++实现如下 vector<vector<int>> generate(int

LeetCode (13) Pascal's Triangle (杨辉三角 )

题目描述 Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5, Return 从第三行开始,每行除了最左边和最右边两个数为1,其他数字都是上一行中相邻两个数字之和.根据上述规则可以写出下面的代码: class Solution { public: vector<vector<int> > generateRow1() { vector<in

118. Pascal's Triangle

Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 分析第k层有k个元素,每层第一个及最后一个元素值为1,对于(k>2)层,第n(n>1 && n < k)个元素A[k][n] = A[k-1][n-1]+A[k-1][n];

119. Pascal's Triangle II

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Note:Could you optimize your algorithm to use only O(k) extra space? 1 class Solution { 2 public: 3 vector<int> getRow(int rowIndex) { 4 5 vect

Java [Leetcode 119]Pascal's Triangle II

题目描述: Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. 解题思路: 每次在上一个list前面插入1,然后后面的每两个间相加赋值给前一个数. 代码描述: public class Solution { public List<Integer> getRow(int rowIndex) { List<Integer> r

LeetCode OJ：Pascal's Triangle（帕斯卡三角）

Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 帕斯卡三角,很简单的问题,见代码: 1 class Solution { 2 public: 3 vector<vector<int>> generate(int numRows) {

Pascal's Triangle II 解答

Question Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. Solution Similar with Pascal's Triangle. Note that the index starts from 0. 1 public class Solution { 2 public List<Integer> getRow(in

[leedcode 119] Pascal's Triangle II

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3,Return [1,3,3,1]. public class Solution { public List<Integer> getRow(int rowIndex) { //由于需要O(k)空间,因此需要借助两个数组保存中间值,并交换两个数组,注意交换的方法! List<Integer> list=new

LeetCode——Pascal's Triangle

Description: Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5, Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] public class Solution { public List<List<Integer>> generate(int numRows) { List