首页 > 代码库 > Bestcoder #21&&hdoj 5139 Formula 【另类打表之分块】
Bestcoder #21&&hdoj 5139 Formula 【另类打表之分块】
Formula
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 155 Accepted Submission(s): 69
Problem Description
You are expected to write a program to calculate f(n) when a certain n is given.
Input
Multi test cases (about 100000), every case contains an integer n in a single line.
Please process to the end of file.
[Technical Specification]
1≤n≤10000000
Please process to the end of file.
[Technical Specification]
Output
For each n,output f(n) in a single line.
Sample Input
2 100
Sample Output
2 148277692
题意:计算i从1~n的i^(n-i-1)的乘积。
普通打表 超内存+超时,,,
不妨将所有的输入数据都存进一个结构体中一个是no(表示输入时的顺序)一个是num(表示输入的数据n),再按照num的大小排序,之后依次找与num相等的数,并将其的值赋给out中
没想到还可以这样。。学习了
代码:
#include <cstdio> #include <cstring> #include <algorithm> #define LL __int64 using namespace std; const LL M = 100005; const LL mod = 1000000007; struct node{ LL no, num; }s[M]; LL out[M]; int cmp(node a, node b){ return a.num < b.num; } int main(){ LL n, cnt = 0; while(scanf("%I64d", &n) == 1){ s[cnt].no = cnt; s[cnt].num = n; cnt++; } LL a, b, index = 0, i; sort(s, s+cnt, cmp); a = 1; b = 1; for(i = 1; index != cnt; i ++){ a = (a*i)%mod; b = (b*a)%mod; while(index !=cnt &&s[index].num == i){ out[s[index++].no] = b; } } for(i = 0; i < cnt; i ++){ printf("%I64d\n", out[i]); } return 0; }
Bestcoder #21&&hdoj 5139 Formula 【另类打表之分块】
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