题目 解题过程: //物理数学题 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { double h,l,v,ans; while(scanf("%lf%lf%lf",&h,&l,&v)!=EOF) { &&l==&&v==)break; ans = h + v *…

NPY and shot Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1035    Accepted Submission(s): 428   Problem Description NPY is going to have a PE test.One of the test subjects is throwing the sh…

NPY and shot Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description NPY is going to have a PE test.One of the test subjects is throwing the shot.The height of NPY is H meters.He can throw the shot at t…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4569 题意:给你一个最高幂为4的一元多项式,让你求出一个x使其结果模p*p为0. 题解:f(x)%(p*p)=0那么一定有f(x)%p=0,f(x)%p=0那么一定有f(x+p)%p=0. 所以我们可以开始从0到p枚举x,当f(x)%p=0,然后再从x到p*p枚举,不过每次都是+p,找到了输出即可,没有的话No solution! AC代码: #include <iostream> #includ…

题目描述:有一个公式,Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, .... n).,如果给出A0, An+1, 和 C1, C2, .....Cn要你计算出A1是多少. 解题报告:一个简单的数学题,我的做法先将公式变形得到:A[n] = 2*A[n-1]+2*C[n-1]-A[n-2],是分别定义两个数组,设A1 = x,然后两个数组里面一个存放第n项的系数,另一个存放第n项的常数,这样一直递推到第n+1项,然后直接解一元一次方程就可以了.代码附上: #inc…

physics 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5826 Description There are n balls on a smooth horizontal straight track. The track can be considered to be a number line. The balls can be considered to be particles with the same mass. At the…

转载链接 因为:Ai=(Ai-1+Ai+1)/2 - Ci,        A1=(A0  +A2  )/2 - C1;       A2=(A1  +  A3)/2 - C2 , ... =>    A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2) =>    A1+A2 =  A0+A3 - 2(C1+C2)  同理可得:       A1+A1 =  A0+A2 - 2(C1)        A1+A2 =  A0+A3 - 2(C1+C2)       A1+A3 =…

NPY and shot Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 808    Accepted Submission(s): 342 Problem Description NPY is going to have a PE test.One of the test subjects is throwing the shot.T…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1030 Delta-wave Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7163    Accepted Submission(s): 2772 Problem Description A triangle field is numbe…

题目链接:pid=4950http://acm.hdu.edu.cn/showproblem.php?pid=4950">http://acm.hdu.edu.cn/showproblem.php?pid=4950 Monster Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 220    Accepted Submission…