UVA 11384 Help is needed for Dexter
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给定序列1...n,每次可以选中序列几个数,然后同时减去相同到一个数,求把序列全部变成0所需要到最小次数.
做法是每次选择序列中最中间到位置到那个数,然后中间到最右边的区间到数全部减去这个值,如 1 2 3 4 5 6 选择4 减掉后得 1 2 3 0 1 2 这个序列其实跟1 2 3 是一样的.
一直这样到操作直到n变成0,复杂度为lgn
#include <iostream>#include <cstdio>using namespace std;int n;int main(){ while(~scanf("%d", &n)){ int res = 0; while(n){ n /= 2; res++; } printf("%d\n", res); } return 0;}
- UVA 11384 - Help is needed for Dexter
- uva 11384 Help is needed for Dexter
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- UVA 11384 - Help is needed for Dexter
- UVA 11384 Help is needed for Dexter
- UVA 11384 - Help is needed for Dexter
- help is needed for dexter uva 11384
- UVa:11384 Help is needed for Dexter
- UVA - 11384 Help is needed for Dexter
- uva 11384 - Help is needed for Dexter
- Uva 11384 - Help is needed for Dexter
- Uva 11384 Help is needed for Dexter
- UVA 11384 Help is needed for Dexter
- uva-11384-Help is needed for Dexter
- UVA - 11384 Help is needed for Dexter
- uva 11384 - Help is needed for Dexter
- UVA 11384 Help is needed for Dexter
- UVA 11384 Help is needed for Dexter
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