vjudge上时限1s 我们按严格的来) 但仔细想了一下 我仍然只发现暴力是不可以的 我暴力的思路是 乘起来 但直接乘会爆 就留个9位吧(感觉有点靠运气) 这样的思路肯定是错的 (像这位小哥) 也不要想去优化 不然就像我搞了半天 仍然TLE 接着 再仔细想想 发现求最后一位非0的数 只需要我们在算的过程中除10 ...
POJ 1150 The Last Non-zero Digit 超大组合数:求超大组合数P(n, m)的最后一个非零位。 4.1更加复杂的数学问题 模运算的世界 今天过节,管它什么节,对我来说都一样,来刷一题渲染一下节日气氛。 终于刷到【登峰造极——高级篇】了,我却一点登峰造极的感觉都没有 P(n, m)=n! / (n-m)!,问题归结...
poj 1150 The Last Non-zero Digit 1 /** 2 大意: 求A(n,m)的结果中从左到右第一个非零数 3 思路: 0是由2*5的得到的,所以将n!中的2,5约掉可得(2的数目比5多,最后再考虑进去即可) 4 那n!中2 的个数怎么求呢? 5 int get2(int n){ 6 if(n==0) 7 return 0; 8 return n/2...
usingnamespacestd; constinttable[5][5]={ {6,2,4,8}, {1,3,9,7}, {1,7,9,3}, {1,9,1,9} }; intget2(intn) { if(n==0)return0; returnn/2+get2(n/2); } intget5(intn) { if(n==0)return0; returnn/5+get5(n/5); } intg(intn,intx) { if(n==0)return0; ...
POJ 1150-The Last Non-zero Digit 题意: 求排列数n,m的最后一个非零数字。 下面是超时程序: #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<cstdio> #include<cmath> using namespace std; int fun(int a, int b, int c) //a的b次方对c取模 { if(b ==...
int last_digit = 1; if(two < five) return 5; else{ if(two > five) last_digit *= table[0][(two-five)%4]; last_digit *= table[1][three%4]; last_digit *= table[2][seven%4]; last_digit *= table[3][nine%4]; return last_digit % 10; ...
ON THE LAST DIGIT AND THE LAST NON-ZERO DIGIT OF nn IN BASE b In this paper we study the sequences defined by the last and the last non-zero digits of nn in base b. For the sequence given by the last digits of nn in b... JM Grau,AM Oller-Marcen - 대한수학회보 ...
DEVICE FOR DISCRIMINATING THE LAST NON-ZERO DIGIT FROM SERIES CODEBOYUN VITALIJ P,SU
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