标签：技术 i++ square desc 无法 贪心 ace ++ while
Goffi and Squary Partition Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1442 Accepted Submission(s): 479 Problem Description Recently, Goffi is interested in squary partition of integers. A set X of k distinct positive integers is called squary partition of n if and only if it satisfies the following conditions: [ol] the sum of k positive integers is equal to n one of the subsets of X containing k?1 numbers sums up to a square of integer. [/ol] For example, a set {1, 5, 6, 10} is a squary partition of 22 because 1 + 5 + 6 + 10 = 22 and 1 + 5 + 10 = 16 = 4 × 4. Goffi wants to know, for some integers n and k, whether there exists a squary partition of n to k distinct positive integers. Input Input contains multiple test cases (less than 10000). For each test case, there‘s one line containing two integers n and k (2≤n≤200000,2≤k≤30). Output For each case, if there exists a squary partition of n to k distinct positive integers, output "YES" in a line. Otherwise, output "NO". Sample Input 2 2 4 2 22 4 Sample Output NO YES YES
题意：输入整数n和k，要求把n分成k个数之和的形式，其中存在k-1个数之和为一个完全平方数，而且这k个数各不相同。
分析： 我们尝试枚举那个完全平方数 S,然后看能否将他拆分为 K-1 个数,并且不用到N-S 这一步可以用贪心+一次调整来搞定。为了保证 K-1 个数都不同,我们尝试尽量
用 1,2,3...这些连续自然数来构造,如果 N-S 出现在这些数中,那么将 N-S 移除,再新加一个数。最后一个数由S-sum（1~k-2）(包括调整过的)来得到。
1.1 前k-2个数中不存在N-S，那么原数列为1，2，3，....，k-2，其中的和大于等于S值，且最小的数为1，没有剩余的空间减少这k-2个数的和
1.2 前k-2个数中存在N-S，设x等于N-S那么原数列为1，2，....x-1，x+1，.....，k-1，其中多出来的空间为避免N-S，同样不存在剩余空间减少和
#include <cstdio> using namespace std; int pnt[35],top; int n,k; bool check(int x) { int sum=0,top=0; int r=n-x,cc=0,cnt=1; pnt[top++]=0; for(int i=0; i<k-2; i++) { cc++; if(cc==r) cc++; pnt[top++]=cc; sum+=cc; } if(sum>=x) return false; pnt[top]=x-sum; if(pnt[top]<=pnt[top-1]) return false; if(pnt[top]==r) { if(pnt[top-1]+1>=pnt[top]-1) return false; } return true; } int main() { while(~scanf("%d%d",&n,&k)) { int flag=0; for(int i=1; i<=2000; i++) { if(i*i>=n) break; if(check(i*i)) { printf("%d\n",i*i); printf("YES\n"); flag=1; break; } } if(flag) continue; printf("NO\n"); } return 0; }
hdu 4982 Goffi and Squary Partition
标签：技术 i++ square desc 无法 贪心 ace ++ while
原文地址：http://www.cnblogs.com/zsyacm666666/p/6048882.html
