B - 尺取 POJ - 2100

最新推荐文章于 2020-07-20 22:12:23 发布

sea_insky

最新推荐文章于 2020-07-20 22:12:23 发布

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本文链接：https://blog.youkuaiyun.com/weixin_44745921/article/details/88564931

本文详细解析了POJ-2100问题，即寻找一个数是否能表示为若干连续正整数平方数的和。通过尺取法实现，算法边生成数字边调整区间，确保总和等于目标值。提供了完整的C语言代码实现。

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B - 尺取 POJ - 2100

ing George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.
After a consultation with his astrologer, King George decided that the lengths of section sides must be a sequence of successive positive integer numbers. A section with side length s contains s 2 graves. George has estimated the total number of graves that will be located on the graveyard and now wants to know all possible graveyard designs satisfying the condition. You were asked to find them.

Input

Input file contains n — the number of graves to be located in the graveyard (1 <= n <= 10 14 ).

Output

On the first line of the output file print k — the number of possible graveyard designs. Next k lines must contain the descriptions of the graveyards. Each line must start with l — the number of sections in the corresponding graveyard, followed by l integers — the lengths of section sides (successive positive integer numbers). Output line’s in descending order of l.

Sample Input

2030
Sample Output
2
4 21 22 23 24
3 25 26 27

Sample Output

2
4 21 22 23 24
3 25 26 27

简单来说就是求一个数是否存在几个连续的平方数之和，尺取法做，边做边生成数字，小了加，大了减。

#include<stdio.h>
#include<math.h>
int main(){
	int m,t,j,a[100000][2],i;
	long long n,sum,L,R;
	while(~scanf("%lld",&n)){
	m=i=sum=0;t=(int)sqrt(n)+1;L=R=1;
	 while(L<=t){
	 	while(R<=t&&sum<n){
	 		sum=sum+R*R;R++;//printf("%lld ",sum);
		 }
		 if(sum==n){
		 	a[m][0]=L;a[m][1]=R;m++;
		 }
		 sum=sum-L*L;L++;//printf("%lld ",sum);
	 	}
	 printf("%d\n",m);
	 for(i=0;i<m;i++){
	 	printf("%d ",a[i][1]-a[i][0]);
	 	for(j=a[i][0];j<a[i][1];j++)
	 	printf("%d ",j);
	 	printf("\n");
	 }}
   return 0;
}