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HDU - 6033 Add More Zero

题意:已知m,求使10k不超过2m-1的最大的k。

分析:

1、2m都是以2,4,6,8结尾,因此减1不影响2m-1的位数。

2、令10k=2m,则k=m * log102,下取整即可。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 10000 + 10;const int MAXT = 10000 + 10;using namespace std;int main(){    int m;    int kase = 0;    while(scanf("%d", &m) == 1){        printf("Case #%d: %d\n", ++kase, (int)(m * log10(2.0)));    }    return 0;}

  

HDU - 6033 Add More Zero