Shrinking Polygons hust 14048

Shrinking Polygons

Time Limit: Unknown

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

 

A polygon is said to be inscribed in a circle when all its vertices lie on that circle. In this problem you will be given a polygon inscribed in a circle, and you must determine the minimum number of vertices that should be removed to transform the given polygon into a regular polygon, i.e., a polygon that is equiangular (all angles are congruent) and equilateral (all edges have the same length).

When you remove a vertex v from a polygon you first remove the vertex and the edges connecting it to its adjacent vertices w₁ and w₂, and then create a new edge connecting w₁ and w₂. Figure (a) below illustrates a polygon inscribed in a circle, with ten vertices, and figure (b) shows a pentagon (regular polygon with five edges) formed by removing five vertices from the polygon in (a).

 

In this problem, we consider that any polygon must have at least three edges.

 

Input

The input contains several test cases. The first line of a test case contains one integer N indicating the number of vertices of the inscribed polygon (3 ≤ N ≤ 10 ⁴). The second line contains N integers X _i separated by single spaces (1 ≤ X _i ≤ 10 ³, for 0 ≤ i ≤ N -1). Each X _i represents the length of the arc defined in the inscribing circle, clockwise, by vertex i and vertex ( i+1) mod N. Remember that an arc is a segment of the circumference of a circle; do not mistake it for a chord, which is a line segment whose endpoints both lie on a circle.

The end of input is indicated by a line containing only one zero.

 

Output

For each test case in the input, your program must print a single line, containing the minimum number of vertices that must be removed from the given polygon to form a regular polygon. If it is not possible to form a regular polygon, the line must contain only the value -1.

Sample input 3 1000 1000 1000 6 1 2 3 1 2 3 3 1 1 2 10 10 40 20 30 30 10 10 50 24 26 0

Output for the sample input 0 2 -1 5

ACM ICPC :: South American Regional 2008

Source

South America 2008-2009

给出n个数字，求最少合并几次能使每一堆的value和相同。

大牛的做法，hash处理sum[i],枚举起点px，每次探测 每一堆的和dd，若sum[end]+t*dd都能hash到，那么这个解就ok。

#include <algorithm> #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const int P = 999997; int n,m; int a[20005]; int hash[1100005]; int Max(int x,int y) { return x>y?x:y; } void insert_hash(int x) { int k = x % P; while (hash[k] >= 0) k++; hash[k] = x; } bool is_in(int x) { int k = x % P; while (hash[k] >= 0 && hash[k] != x) k++; if (hash[k] == x) return true; else return false; } int main() { int i,j,k,s,t,ans,dd,px; int x,y; while (cin >>n) { if (n == 0) break; memset(hash,-1,sizeof(hash)); memset(a,0,sizeof(a)); for (i=1;i<=n;i++) { cin >>a[i]; a[i] += a[i-1]; insert_hash(a[i]); //printf("%d ",a[i]); } m = n*2; ans = 0; for (i=n+1;i<=m;i++){ a[i] = a[n] + a[i-n]; insert_hash(a[i]);//printf("%d ",a[i]); } for (px=1;px<=n;px++) { bool bo = false; for (i=px;i<px+n-2;i++) { dd = a[i] - a[px-1]; if (a[n] % dd == 0) { k = a[i]; while (k < a[px+n-1]) { if (!is_in(k)) break; k += dd; } if (k == a[px+n-1]) {bo = true; break;} } if (a[n]/dd <= ans) break; } if (bo && a[n]/dd > ans) {ans = a[n]/dd;} } if (ans < 3) cout <<"-1"<<endl; else cout <<n - ans<<endl; } return 0; }

这样还是有点耗时，更省力的解法是将sum[i]%dd存在数组B[]中，查找B[i]中出现次数最多的元素(sort一下做),次数记为maxn.if(maxn==分的组数)，则ok，这样分组数从N到3枚举就可以了。