# HDU 3641-Treasure Hunting-分治-[解题报告]HOJ

2016-03-14 14:58:22来源:[db:出处]作者:[db:作者]人点击

Treasure Hunting

Zstu_yhr is a very curious person who fell in love with math when he was in elementary school phase. When he entered the middle school, he learned Multiplication and Power Multiplication. yhr is so ambitious that he not only dreams to be a mathematician but also dreams to be richer than Bill Gates.One day, he is suddenly encountered with a crazy thought that is to hunt treasure to make one of his dreams a reality. Since yhr is such a strong-willed person that he will never give up as long as his goal has not been achieved. After going through 9*9 challenges, as a reward of god for that hard, he finally discovers an antique hole which is very likely to have a good number of treasures in it. However, as every novel writes, he can never get the treasures so easily. He has to open a coded door at first. He finds that there are 2*N numbers on the door. He speculates that they must be able to generate the password. Disappointedly, there isn’t any clue left for him. He has no better way but to YY. Firstly, he divides these 2*N number into N piles equally. The first pile is composed of a1,b1 and the second pile is composed of a2,b2…certainly, the i-th pile is composed of ai,bi…After completing this task, he calculates a1^b1*a2^b2*a3^b3…*an^bn and gets its result M. He takes M as the password to open the door. What’s a pity, he fails. Then he starts to YY again. Maybe the right password is the minimum number x which satisfies the equation x!%M=0. So he wants to have a try. But he doesn’t know how to get the number so that he has to turn to you for help. Can you help him?

In the first line is an integer T (1<=T<=50) indicating the number of test cases.Each test case begins with an integer n (1<=n<=100), then followed n lines. Each line contains two numbers ai and bi (1 <= ai <= 100, 1<=bi<=10000000000000)

In the first line is an integer T (1<=T<=50) indicating the number of test cases.Each test case begins with an integer n (1<=n<=100), then followed n lines. Each line contains two numbers ai and bi (1 <= ai <= 100, 1<=bi<=10000000000000)

123 24 1

6Hint n! is the factorial of number n:0!=1n!=n*(n-1)! (n>=1)a^0=1 (a>=1)a^i=a*(a^i-1) (i>=1)

/*Pro: 0Sol:date:*/#include #include #include #define inf 99999999999999999999LLusing namespace std;int prime,sub;__int64 num;void getprime(){ bool flag ; sub = 0; memset(flag , true, sizeof(flag)); for(int i = 2; i <= 110; i ++) for(int j = i + i; j <= 110;j += i) flag[j] = false; for(int i = 2; i <= 100; i ++) if(flag[i]) prime[sub ++] = i;}__int64 getnum(__int64 x, int p){// __int64 sum = 0; while(x % p == 0){ sum ++; x /= p; }return sum;}bool can(__int64 indx){ for(int i = 0; i < sub; i ++){// printf("%d %d/n",i,prime[i]); 我还以为又是数组越界的奇怪问题呢，原来是这个函数被多次调用 if(prime[i] && num[prime[i]]){ __int64 sum = 0; for(__int64 k = prime[i]; k <= indx; k *= prime[i]) sum += indx / k; if(sum < num[prime[i]]) return false; } } return true;}__int64 bin(){ __int64 low = 0, high = inf, mid ; while(low <= high){ mid = (low + high) >> 1;// printf("%d&&/n",mid); if(can(mid)) high = mid - 1; else low = mid + 1; } return low;}int main(){ getprime();// __int64 kk = 10000000000000 * 100; __int64 sum = 0;// 99999999999999999999// __int64 in = 33333333333333333333;// for(__int64 ka = 97; ka <= in; ka *= 97)// sum += in / ka;// if(sum >= kk) puts("yes");// else puts("no"); int t,n,a; __int64 b; scanf("%d",&t); while(t -- ){ scanf("%d",&n); memset(num,0,sizeof(num)); for(int i = 0; i < n; i ++){ scanf("%d%I64d",&a,&b); for(int j = 0; j < sub; j ++){ if(a % prime[j] == 0){ num[prime[j]] += getnum(a,prime[j]) * b; } } } printf("%I64d/n",bin()); } return 0;} 