The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5706    Accepted Submission(s): 2311 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5956    Accepted Submission(s): 2427 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

母函数的特殊情况,左右两边都可以放,如样例1,2,9 母函数为(1+x+1/x)*(1+x^2+1/x^2)*(1+x^9+1/x^9) 化简为(1+x+x^2)*(1+x^2+x^4)*(1+x^9+x^18)/(x*x^2*x^9) 这样就好计算了,看代码: ;}…

题意: 有一个天平.有N个砝码.重量分别是A1...AN. 问重量[1..S]中有多少种重量是无法利用这个天平和这些砝码称出来的. S是N个砝码的重量总和. 思路: 对于每一个砝码来说,有三种:不放,放左盘,放右盘. 额,,母函数和DP其实核心一样,,,, 看代码,, 代码: int n; int aa[105]; bool a[10005], b[10005]; int temp[10005]; int main(){ while(scanf("%d",&n)!=EOF){…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6652    Accepted Submission(s): 2730 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

题意: 现在你被要求用天平和一些砝码来量一剂药.当然,这并不总是可以做到的.所以你应该找出那些不能从范围[1,S]中测量出来的品质.S是所有重量的总质量. 输入一个n,后面有n个数,表示这n个物品的质量 题解: 注意这个题的题干,这个天平只要能制造出来那个质量差,然后这个质量差就不会输出(看懂第二个样例就行) 这道题是母函数的变化版,没学过母函数的可以看一下:母函数 <普通母函数(HDU - 1028 ) && 指数型母函数(hdu1521)> 代码: 1 #include&l…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7677    Accepted Submission(s): 3187 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4497    Accepted Submission(s): 1808 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

母函数,指数可以为1也可以为-1,扩大指数加消减发现TLE,于是采用绝对值就过了. #include <stdio.h> #include <string.h> #define MAXNUM 10001 int c1[MAXNUM], c2[MAXNUM]; ]; int myabs(int x) { ? -x:x; } int main() { int n, sum, tmp; int i, j, k; while (scanf("%d", &n) !…

MB,一开始就想到是不是只要加上一个不选择砝码的情况,但一直没动手做,因为看了看网上了,觉得总有点复杂,认为自己想错了.... 相信自己 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define N 101 using namespace std; int c1[N*N],c2[N*N]; int Fa[N],ans[N*N],ap; int m…