http://blog.csdn.net/huxin2007/archive/2006/09/28/1302156.aspx
遞歸
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斐波納契數(遞歸實現)不可行
...{
if(i < 1) return 0;
if(i == 1) return 1;
return F(i-1) + F(i-2);
}
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斐波納契數(動態規劃)
...{
int t;
if(knownF[i] != unknown) return knownF[i];
if(i == 0) t = 0;
if(i == 1) t = 1;
if(i > 1) t = F(i-1) + F(i-2);
return (knownF[i] = t);
}
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揹包問題(遞歸實現)不可行
...{
int size;
int val;
} Item;
Item items[N];
int knap(int cap)
...{
int i, space, max, t;
for(i = 0, max = 0; i < N; i++)
...{
if((space = cap - items[i].size) >= 0)
if((t = knap(space) + items[i].val) > max)
max = t;
}
return max;
}
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揹包問題(動態規劃)
...{
int i, space, max, maxi, t;
if(maxKnown[cap] != unknown) return maxKnown[cap];
for(i = 0, max = 0; i < N; i++)
...{
if((space = cap - items[i].size) >= 0)
...{
if((t = knap(space) + items[i].val) > max)
...{
max = t;
maxi = i;
}
}
}
maxKnown[cap] = max;
return max;
}