题目大意是一个人有g次猜数机会和l条生命线,每次猜数如果猜错了则扣除一次猜数机会,如果猜高了还要扣除一条生命线,如果在没有生命线的情况下猜高了就输了。问在有g次猜数机会和l条生命线的情况下最多能保证正确猜出多少以内的数?
用f[i,j]表示g=i,l=j的情况下可以保证猜中的最大数,将每次猜数分成三种情况,猜低、猜中和猜高,则可推出状态转移方程f[i,j]=f[i-1,j]+1+f[i-1,j-1];由于l=0的情况下只能猜低的数,因此初始化f[i,0]=i;
1651819 2008-09-29 22:39:48 Accepted 1503 FPC 0 116 IwfWcf
chenlina20160429
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