Raising Modulo Numbers(ZOJ 2150)】的更多相关文章

这题其实就是快速求一个高次幂的模. 这是题目的答案 #include<iostream> #include<cmath> using namespace std; ]; ]; int mod(int a, int b, int c) { ; while(b) { ) z = (z*a)%c; b/=; a = (a*a)%c; } return z; } int main() { int T; cin>>T; while(T--) { int M; int H; cin…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6373   Accepted: 3760 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5477   Accepted: 3173 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
POJ1995 Raising Modulo Numbers 计算(A1B1+A2B2+ ... +AHBH)mod M. 快速幂,套模板 /* * Created: 2016年03月30日 23时01分45秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5934   Accepted: 3461 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6347   Accepted: 3740 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5500   Accepted: 3185 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
-->Raising Modulo Numbers Descriptions: 题目一大堆,真没什么用,大致题意 Z M H A1  B1 A2  B2 A3  B3 ......... AH  BH 有Z组数据   求(A1B1+A2B2+ ... +AHBH)mod M. Sample Input 3 16 4 2 3 3 4 4 5 5 6 36123 1 2374859 3029382 17 1 3 18132 Sample Output 2 13195 13 题目链接https://v…
Raising Modulo Numbers Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5532   Accepted: 3210 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that…
People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market…
People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market…
阶乘总和 题目大意:要你算一堆阶乘对m的模... 大水题,对指数二分就可以了... #include <iostream> #include <functional> #include <algorithm> using namespace std; typedef long long LL_INT; LL_INT cal(const LL_INT, const LL_INT, const LL_INT); int main(void) { int case_sum,…
快速幂取模 #include<cstdio> int mod_exp(int a, int b, int c) { int res, t; res = % c; t = a % c; while (b) { ) res = res * t % c; t = t * t % c; b >>= ; } return res; } int main() { int T; scanf("%d",&T); while(T--) { int m,h; scanf(&…
ZOJ2150 快速幂,但是用递归式的好像会栈溢出. #include<cstdio> #include<cstdlib> #include<iostream> #include<cmath> using namespace std; long long M,i; #define LL long long int _work(LL a,LL n) { LL ans=1; while(n){ if(n&1){ ans=(ans*a)%M; n--; }…
题意: 思路: 对于每个幂次方,将幂指数的二进制形式表示,从右到左移位,每次底数自乘,循环内每步取模. #include <cstdio> typedef long long LL; LL Ksm(LL a, LL b, LL p) { LL ans = 1; while(b) { if(b & 1) { ans = (ans * a) % p; } a = (a * a) % p; b >>= 1; } return ans; } int main() { LL p, a…
题目:http://poj.org/problem?id=1995 题目解析:求(A1B1+A2B2+ ... +AHBH)mod M. 大水题. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; int n,mod,sum; int main() { ],b[…
二进制前置技能:https://www.cnblogs.com/AKMer/p/9698694.html 题目传送门:http://poj.org/problem?id=1995 题目就是求\(\sum_{i=1}^na[i]^{b[i]}mod\) \(m\).我们只要会快速求\(a^b\)就行了. 我们可以用二进制拆分思想,把\(a^b\)转化成\(a^{(1010...1)_2}\)之类的.然后根据\(a^{x+y}=a^x*a^y\),我们可以将\(a^b\)转化成\(a^{(10000…
[题目链接] http://poj.org/problem?id=1995 [算法] 快速幂 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #includ…
题目链接:http://poj.org/problem?id=1995 解题思路:用整数快速幂算法算出每一个 Ai^Bi,然后依次相加取模即可. #include<stdio.h> long long quick_mod(long long a,long long b,long long c) { long long ans=1; while(b) { if(b&1) { ans=ans*a%c; } b>>=1; a=a*a%c; } return ans; } int…
[题目链接] http://poj.org/problem?id=1995 [算法] 基本快速幂(二进制思想) 注意两个int相乘可能溢出,加(long long)但是相乘不要加括号,不然会先溢出在类型转换 #include <iostream> using namespace std; int z,m,h,cur,ans,a,b; void calc() { cin>>a>>b; cur = % m; ) { ) cur = (long long)cur * a %…
嗯... 题目链接:http://poj.org/problem?id=1995 快速幂模板... AC代码: #include<cstdio> #include<iostream> using namespace std; int main(){ ; scanf("%lld", &N); while(N--){ scanf("%lld%lld", &M, &n); sum = ; ; i <= n; i++){…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 9512 Accepted: 5783 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others…
Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that…
Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that…
题目链接: https://vjudge.net/problem/POJ-1995 题目大意: 求一堆ab的和模上m 思路: 直接上模板 #include<iostream> #include<vector> #include<queue> #include<algorithm> #include<cstring> #include<cstdio> #include<set> #include<cmath> u…
#include <cstdio> typedef long long ll; int quick_pow(ll a,ll b,ll mod){ ll ans=; ))ans=(ans*a)%mod; return ans; } int main(){ int z,m,h,a,b,ans; for(scanf("%d",&z);z--;){ scanf(; while(h--)scanf("%d%d",&a,&b),ans=(an…
1. poj 1995  Raising Modulo Numbers 2.链接:http://poj.org/problem?id=1995 3.总结:今天七夕,来发水题纪念一下...入ACM这个坑也快一年了 题意:求ai^bi和模m.裸快速幂 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<…
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