if(a[start] + T == a[j]) {

                        start++;

                    }else{

                        start =1;

                        T = a[j] - a[start];

                        start++;

                    }

如果在start>2时，进入else分支，那么至少a[j-1]-a[start-1]验证过不是周期，但a[j-1]-a[1]肯定没验证到