一道有挑战性的题

原题目链接：http://blog.youkuaiyun.com/masterluo/article/details/4963960

从1~N这N个数按照下面的排法一直排下去，规定S(N)为左上角是1，右下角为N的这个大矩形里面的所有小矩形里的数的和。例如S(12)就表示图中12个绿色格子的数的和。现在给定N，让你求S(N)。

#include "stdio.h"

#define MX 10
long sum[MX+1];
int dim[MX+1];

void cal(){
	int i, j;
	for(i=0; i<=MX; i++) 
		dim[i] = i*i;
	sum[0] = 0;
	for(i=1; i<=MX; i++){
		sum[i] = sum[i-1];
		for(j=dim[i-1]+1; j<=dim[i]; j++)
			sum[i] += j;
	}
}

void main(){
	int u;
	int lev, i, j, count;
	int start, end;
	int mark;
	int sums;

	cal();
	
	while(printf("输入："), scanf("%d", &u), u){

		for(lev=1; lev<=MX; lev++){
			if(u<=dim[lev])
				break;
		}
		
		if(lev>MX){
			printf("数值太大！\n----------------\n");
			continue;
		}

		mark = 1;
		sums = sum[lev];

		if(dim[lev]-lev<u){
			count=u-dim[lev-1]-1;
			for(i=lev-(dim[lev]-u+1); i>0; i--){
				if(mark){
					start = dim[lev-1]+1;
					for(j=0; j<count; j++)
						sums -= start+j;
					mark = 0;
				}else{
					end = dim[lev];
					for(j=0; j<count; j++)
						sums -= end-j;
					mark = 1;
				}
				count -= 2;
				lev--;
			}
		}else{
			count = dim[lev]-u;
			for(i=lev-(u-dim[lev-1]); i>0; i--){
				if(mark){
					end = dim[lev];
					for(j=0; j<count; j++)
						sums -= end-j;
					mark = 0;
				}else{
					start = dim[lev-1]+1;
					for(j=0; j<count; j++)
						sums -= start+j;
					mark = 1;
				}
				count -= 2;
				lev--;
			}
		}

		printf("结果：%d\n----------------\n", sums);
	}
}

好像是对的……