ZOJ-3706-Break Standard Weight
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Break Standard Weight
With some standard weights, we can measure several special masses object exactly, whose weight are also positive integers in mass units. For example, with two standard weights 1 and 5, we can measure the object with mass 1, 4, 5 or 6 exactly.
In the beginning of this problem, there are 2 standard weights, which masses are x and y. You have to choose a standard weight to break it into 2 parts, whose weights are also positive integers in mass units. We assume that there is no mass lost. For example, the origin standard weights are 4 and 9, if you break the second one into 4 and 5, you could measure 7 special masses, which are 1, 3, 4, 5, 8, 9, 13. While if you break the first one into 1 and 3, you could measure 13 special masses, which are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13! Your task is to find out the maximum number of possible special masses.
Input
There are multiple test cases. The first line of input is an integer T < 500 indicating the number of test cases. Each test case contains 2 integers x and y. 2 ≤ x, y ≤ 100
Output
For each test case, output the maximum number of possible special masses.
Sample Input
2
4 9
13
Time Limit: 2 Seconds Memory Limit: 65536 KB
With some standard weights, we can measure several special masses object exactly, whose weight are also positive integers in mass units. For example, with two standard weights 1 and 5, we can measure the object with mass 1, 4, 5 or 6 exactly.
In the beginning of this problem, there are 2 standard weights, which masses are x and y. You have to choose a standard weight to break it into 2 parts, whose weights are also positive integers in mass units. We assume that there is no mass lost. For example, the origin standard weights are 4 and 9, if you break the second one into 4 and 5, you could measure 7 special masses, which are 1, 3, 4, 5, 8, 9, 13. While if you break the first one into 1 and 3, you could measure 13 special masses, which are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13! Your task is to find out the maximum number of possible special masses.
Input
There are multiple test cases. The first line of input is an integer T < 500 indicating the number of test cases. Each test case contains 2 integers x and y. 2 ≤ x, y ≤ 100
Output
For each test case, output the maximum number of possible special masses.
Sample Input
2
4 9
10 10
13
9
题目链接：Break Standard Weight
题意：给定两个数，你可以把其中一个数拆成两个数（总和不变），然后将这三个数进行加减操作，问可以得到多少种不同的正整数值。
思路：每个数都有选与不选，被+或者被-，三个数也就是(2^2)^3种状态，然后分别枚举x,y可以拆成的两个数。
这里我将一个6位的2进制数，后面3位表示这3个数选与不选，前3位分别表示这3个数是+还是-。
代码：
#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <set>#include <cstring>#include <string>#include <algorithm>#define lson l, m, rt<<1#define rson m+1, r, rt<<1 |1#define lowbit(x) x&(-x)#define LL long longusing namespace std;set<int> sat;int a[5];void fun(int x){ int num = 0; if(x & 1) { num += ((x&32)?1:-1 ) * a[0]; } if(x & 2) { num += ((x&16)?1:-1 ) * a[1]; } if(x & 4) { num += ((x&8)?1:-1 ) * a[2]; } if(num > 0 && !sat.count(num)) { sat.insert(num); }}int main(){ int T, t[3], ans; scanf("%d", &T); while(T--) { scanf("%d%d", &t[0], &t[1]); ans = 0; for(int k = 0;k <= 1;k++) { a[0] = t[k]; for(int i = 1;i < t[k^1];i++) { a[1] = i; a[2] = t[k^1]-i; sat.clear(); for(int j = 0;j < (1<<6);j++) { fun(j); } ans = max(ans,(int) sat.size()); } } printf("%d\n", ans); } return 0;}
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