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本帖最后由 jason680 于 2018-01-29 13:46 编辑
找出"更快"的算法/方法...
money=0
5 2 1
------
x 0 0
count = 1
money=1
5 2 1
------
x 0 1
count = 1
money=2
5 2 1
------
x 0 2
x 1 0
count = 2
money=3
5 2 1
------
x 0 3
x 1 1
count = 2
money=4
5 2 1
------
x 0 4
x 1 2
x 2 0
count = 3
money=5
5 2 1
------
x 0 5
x 1 3
x 2 1
count = 3
...
money 0,1,2,3,4,5,...
count 1,1,2,2,3,3,....
// get_coin_cyc1.c
if(--num == 1) return(money/2+1);
// get_coin_cyc.c
if(--num == 0) return 1;
Note: in coin array
... [3] [2] [1] [0] <== num
... 10 5 2 1 <== value
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