https://vjudge.net/problem/UVA-11040 找规律 #include<cstdio> using namespace std; ][]; int main() { int T,d; scanf("%d",&T); while(T--) { ;i<=;i++) ;j<=i;j++) scanf(-][j*-]); ;i<=;i+=) { ;j<=;j+=) { a[i][j]=a[i-][j-]-a[i][j-]-…

砖块上的数字最终都可以看作是最后一行的线性组合,独立变元最多9个. 这类题的一般做法,线性组合都可以列出方程然后高斯消元. 对于这道题,只要确定最后一行剩下的4个变量就好了,对于最后一行的j位置,它对上面位置某个数字的和贡献次数 等于它到那个位置路径的方案数,可以发现就是杨辉三角.倒数第二行的数已经足够确定剩下的变量,x到对应位置y的方案是2. x = (y - xleft - xright)/2. /***********************************************…

题意:给定一个金字塔,除了最后一行,每个数都等于支撑它的两个数的和,现在给奇数行的左数奇数位置,求整个金字塔. 析:很容易看出来,从下往上奇数行等于 a[i][j] = (a[i-2][j-1] - a[i][j-1] - a[i][j+1]) / 2;然后偶数行就推出来了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string&…

题意: 45块石头如图排列,每块石头上的数等于下面支撑它的两数之和,求其余未表示的数. 分析: 首先来计算最下面一行的数,A71 = A81 + A82 = A91 + 2A92 + A93,变形得到A92 = (A71 - A91 - A93) / 2. 以此类推,就能得到最下面一整行的数.有了这个“地基”以后,所有的数就都能算出来了. #include <cstdio> ][]; int main() { //freopen("in.txt", "r"…

题目链接:XORwice 题意:给你两个数a.b.求一个数x,使得((a异或x)+(b异或x))这个值最小,输出最小那个x 题解: 输出(a|b)-(a&b)就行(猜了一手 代码: #include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> #include<queue> #include<vector> #define mem(x)…

Yup!! The problem name reects your task; just add a set of numbers. But you may feel yourselvescondescended, to write a C/C++ program just to add a set of numbers. Such a problem will simplyquestion your erudition. So, lets add some avor of ingenuity…

A pair of numbers has a unique LCM but a single number can be the LCM of more than one possiblepairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a given positive integer N, thenumber of different integer pairs with LCM is equal to N…

题目链接: 题目 Add All Time Limit:3000MS Memory Limit:0KB 问题描述 Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply…

Add AgainInput: Standard Input Output: Standard Output Summation of sequence of integers is always a common problem in Computer Science. Rather than computing blindly, some intelligent techniques make the task simpler. Here you have to find the summa…

题意  求把全部数加起来的最小代价  a+b的代价为(a+b) 越先运算的数  要被加的次数越多  所以每次相加的两个数都应该是剩下序列中最小的数  然后结果要放到序列中  也就是优先队列了 #include<cstdio> #include<queue> using namespace std; priority_queue<int, vector<int>, greater<int> >q; typedef long long ll; ll…