考虑到以下要求,设计一个公式来找到下面给出的x和y
given: length1 length2, assume length1 >= length2
total = length1 + length2
i = 0 : x = 0, y = 0
i = 1 : x = 0, y = 1
...
i = length2 -1 : x = 0, y = length2 -1
i = total-length1 : x = 0, y = 0
i = total-length1 +1 : x = 1, y = 0
...
i = length1 + length2: x = length1 -1, y = 0
所以在代码中,它看起来像:
int length1 = //given
int length2 = //given
int total = length1 + length2;
for (int i = 0; i < total; i++) {
x = ? //answer here
y = ? //answer here
}
以下是length1 = 5时的示例; length2 = 4
i x,y
---------
i=0 0,0
i=1 0,1
i=2 0,2
i=3 0,3
i=4 0,0
i=5 1,0
i=6 2,0
i=7 3,0
i=8 4,0
编辑: 我正在寻找一个用于寻找x和y的单线 当i小于length2而y为0时,将x out分为0的东西,当i为&gt;长度1。
答案 0 :(得分:1)
if (i < length2) {
x = 0;
y = i;
} else {
x = i - length2;
y = 0;
}
答案 1 :(得分:0)
怎么样:
if (i < length2) {
x = 0;
y = i;
} else {
x = i - length2;
y = 0;
}