http://acm.hdu.edu.cn/showproblem.php?pid=4282

題解：http://www.cnblogs.com/E-star/archive/2012/09/11/2680992.html

View Code #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string>#define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b)#define ll __int64#define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 100007 #define N 107 using namespace std; //freopen("din.txt","r",stdin); ll inf = 2147483648ll;ll Pow(int a,int b){ll sum = 1;for (int i = 1; i <= b; ++i) sum *= a;return sum; } bool Bsearch(int L,int R,int x,int z,int k,ll POW){int l = L, r = R;while (l <= r){int m = (l + r)>>1;ll tmp = POW + Pow(m,z) + (ll)x*m*z;if (tmp < 0) r = m -1;//注意這里處理ll溢出時的操作wa好幾次else if (tmp == k) return true;else if (tmp < k) l = m + 1;else r = m - 1;}return false; } int main(){//freopen("din.txt","r",stdin);int K;while (~scanf("%d",&K)){if (!K) break;int ans = 0;for (int x = 1; x < 50000 && x < K; ++x){//枚舉xfor (int z = 2; z < 31; ++z){//枚舉Zll tmp = Pow(x,z);if (tmp > K) break;int mk = Bsearch(x + 1,50000,x,z,K,tmp);//二分查找是否存在滿足的y值if (mk) ans++;}}printf("%d\n",ans);} }

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轉載于:https://www.cnblogs.com/E-star/archive/2012/09/11/2680966.html

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