一些代碼-數(shù)學問題

UVAOJ10061nb0。Log10(a*b)=log10(a)+log(b)0N!分解為素數(shù)的積，記錄下<=B的素數(shù)因子的個數(shù)（B的原因代碼中解釋了N!的因子中（BN!B的因子用來B時結(jié)束B0.usingnamespacestd;intsu[ intdigit(intnn,int{intdoublesum=0;}intzero(intnn,int{inti,j;intintnum;{{{}}}{{{elsegotoend;}}}return}int{intnn,bb;inti;intzz,dd;{cout<<zz<<""<<dd<<endl;}return}:超長方體(1+2+…+n)^4-超立方體usingnamespacestd;intmain(){longlonglonglongs2[101],s3[101],s4[101];longlongr2[101],r3[101],r4[101];{r2[i]=(i*(i+1)/2)*(i*(i+1)/2)-r3[i]=(i*(i+1)/2)*(i*(i+1)/2)*(i*(i+1)/2)-}{cout<<s2[n]<<""<<r2[n]<<""<<r3[n]<<""<<s4[n]<<"}return}UVAOJ10014a0,an+1,c1,c2,…cn,ai=(a(i-1)+a(i+1))/2-ci,a1。先全部累加：a1+an=a0+a(n+1)–2Σci(1<=i<=n)按照上面得出的公式再累加:(n+1)a1n*a0a(n+12ΣΣcj(1<=j<=i#include<stdio.h>#include<string.h>intmain(){intnCases;intnN;doublea0,an1;doublea1;doubleci[3000+10];doublec;doublesum;scanf("%d",&nCases);while(nCases--){scanf("%d",&nN);scanf("%lf%lf",&a0,&an1);sum=0.0;memset(ci,0,sizeof(ci));for(intj=1;j<=nN;++j){scanf("%lf",ci[j]ci[j1]c;//ci[j]代表數(shù)列ci中第1-j的sum+=}a1=(nN*a0+an1-2*sum)/(nN+printf("%.2f",if(nCases){}}return}HDU486120141k，pk1^i+2^i+...+(p-1)^imodp（1i<=k)，現(xiàn)在兩人輪流取球，最后球的值總和大的人贏，問先手是否能贏#include<cstdio>#include#include<cstdio>#include<cstring>#include<cmath>longlongk,intmain()while(~scanf("%lld%lld",&k,&p)){if(k/(p-1)%2)printf("YES\n");elseprintf("NO\n");}return}≡1（moda是整數(shù)，pa,pa的(p-1)p的余1。φ函數(shù)的值通式：φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn),p1,p2……pnx的所有質(zhì)因數(shù)，x0的整數(shù)。φ(1)=1（1互質(zhì)的數(shù)(1)1本身。(12=2*2*3那么φ（12）=12*（1-1/2）*(1-1/3)=4nnφ(n)nn素的正整數(shù)的個數(shù)，稱為n的函數(shù)值，這里函數(shù)φ：N→N，n→φ(n)稱為函數(shù)。m,n互質(zhì)，φ(mn)=φ(m)φ(n)。n為奇數(shù)時，φ(2n)=φ(n),PrimitiveRoot假設一個數(shù)gPg^imodP的結(jié)果兩兩不同,且有1<g<P0<i<P,g可以稱為是P的一個原根,g^(P-11modP)當且僅當指數(shù)P-1的時候成立.(P是素數(shù)).簡單來說，g^imodpg^jmodp（p為素數(shù)sum次操作的話，其實有很多牌是經(jīng)過了多次翻11的總數(shù)的奇偶性是固定的。數(shù)公式必定有除法C（n,m）=n!/(m!*(n-m)!)但是我們知道費定理a^(p-1)=1%p那么a^(p-1)/a=1/a%p得到a^(p-2)=#include<iostream>#include<stdio.h>#include<string.h>#include#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>usingnamespacestd;#definemod#define #definemaxn100000+5LLf[maxn];void{inti;f[0]=1;for(i=1;i<maxn;i++)f[i]=(f[i-1]*i)%mod;}LLquickmod(LLa,LL{LLans=1;{{ans=b--}a=}return}int{intn,m,i,j,k,l,r,x,ll,rr;{l=r=for(i=0;i<n;{if(l>=xlll-x;//當最小的1個數(shù)大于x，把x1elseif(r>=xlll%2)==(x%2))?0:1;//當l<x<=r，由于無論怎么翻，其奇偶性必定相等，所以看l的奇偶性與x是否相同，相同那么知道最小必定變?yōu)?，否則變?yōu)?elsellx-r;//當x>r,那么在把1全部變?yōu)?的同時，還有x-r0變?yōu)閕f(r+x<=mrrr+x;//當r+x<=m的情況下，全部變?yōu)閑lseif(l+x<=mrrl+x)%2m%2)?m:m-1);//r+x>m但是elserr2*m-(l+x);//在l+x>m的情況下，等于我首先把m1變?yōu)榱?，那么我還要翻（l+x-m）張，所以最終得到m-(l+x-m)1l=ll,r=}LLsum=for(i=l;i<=r;i+=2)//使用 }return}HDU49192014 n最大是先乘不部拿出1就是外面加上個kk異或0=k所以拿出外面也k。JAVA方便，上面的遞推式用簡單的遞歸就行了。importimportjava.util.*;importjava.io.*;importpublicclassMainpublicstaticBigIntegertwo=BigInteger.valueOf(2);publicstaticBigIntegerfour=BigInteger.valueOf(4);publicstaticBigIntegersix=BigInteger.valueOf(6);publicstaticHashMap<BigInteger,BigInteger>map=newHashMap<BigInteger,publicstaticBigIntegerF(BigIntegern){if(map.containsKey(n))returnBigIntegerBigIntegertmp1=n.divide(two);BigIntegertmp2=n.mod(two);BigIntegertmp3=tmp1.subtract(BigInteger.ONE);if( pareTo(BigInteger.ONE)==0)tmp1=F(tmp1).multiply(four).add(tmp1.multiply(six));tmp1=map.put(n,tmp1);returntmp1;}publicstaticvoidmain(String[]args){Scannercin=newScanner(System.in);map.put(BigInteger.ZERO,BigInteger.ZERO);map.put(BigInteger.ONE,BigInteger.ZERO);while(cin.hasNext()){BigIntegern=cin.nextBigInteger();}}}HDU2058找一個子序列的和便是Ma+1a+2...數(shù)學題，等差數(shù)列的運用。Sna1+an)*n2a1a1n-1*d)*n/2#include<stdio.h>#include<math.h>int{intn,m,#include<stdio.h>#include<math.h>int{intn,m,a,while(scanf("%d%d",&n,&m)&&(n||{len=(int)sqrt(2*m);while(len--){a=m/(len+1)-len/if((2*a+len)*(len+1)/2==m)printf("[%d,%d]\n",a,}}return}HDU49232014AB數(shù)列(0<=bi<=1)A0...01...1bi=ai0i<jbi<=bjbibjbibj的值都可以得到更優(yōu)的結(jié)果。所以，bi=bj。bi值一定相同。這樣，我們可以得到一個中間序列，x1x1,x2,...,x2,xm,...xm.x1xmxixixibi相等時取得最優(yōu)值。因此，我們可以把這兩段合并，求出B序列一定是這樣：y1y1ym,ym。所以我們可以用一個單調(diào)棧來維bi值。#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<functional>#include<time.h>usingnamespacetypedefpair<double,int>constintINF=constintMAXN=(int)intinta[MAXN];doubleb[MAXN];PDIsk[MAXN//inttail//棧頂指voidvoidsolve()tail0//for(inti=0;i<n;i++)sk[tailmake_pair(1.0*a[i1//whiletail>1&&sk[tail-1].first<=sk[tail-2].first//如果棧不滿足單調(diào)intcnt=sk[tail-1].second+sk[tail-2].second;doubletmp=sk[tail-2]=make_pair(tmp,cnt);}}//for(inti=0,j=0;i<tail;for(intk=0;k<sk[i].second;k++)b[j++]=sk[i].first;doubleans=for(inti=0;i<n;ans+=(a[i]-b[i])*(a[i]-printf("%f\n",}intmain(){intT;scanf("%d",while(T--)scanf("%d",for(inti=0;i<n;i++)scanf("%d",&a[i]);}return}HDU49372014nbase3，4，5，6組成的話，basenn，求有多少個幸運進制。無窮多個的話輸出-1，單個位置9用相應的字符表示。3100001e124位，直接枚舉每一位計算N即可#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<utility>#include<stack>#include<queue>#include<map>#include<deque>#include<cmath>#definemax(x,y)#definemin(x,y)usingnamespace#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<utility>#include<stack>#include<queue>#include<map>#include<deque>#include<cmath>#definemax(x,y)#definemin(x,y)usingnamespacelonglongx,N;intans;inta[6];map<int,bool>mapp;boolcheck(longlongx)//{{if(x>=3&&x<=6)return1;elsereturn0;}for(inti=3i<=min(6,N-1i++)//6N-1if((x-i)%N==0&&check((x-i)/N))//3-6能被三整出，returnreturn}voiddfs(intp)//100003，4，5，6枚舉這三位二分進制算{if(p==3||{longlongl=1e4,r=1e12;//最大最小邊界if(p==4r=1e6;//如果四位那么最大邊界可以縮小{longlongm=(l+r)/2;longlongtmp=0;for(inti=1;i<=p-1;i++)if(tmp>x)r=m;elsel=m;}longlongfor(inti=1;i<=p-1;tmp=tmp*l+a[i];//dfs出來的結(jié)果來算xif(tmp==x&&l!=1e4&&mapp[l])//10000map中之前沒有這進制{}if(l!=r&&r!=1e4)//r{for(inti=1;i<=p-1;i++)if(tmp==x&&{}}}if(p==4)for(inti=3i<=6i++)//3-6{}}int{inttt;for(intfor(intt=1;t<=tt;{printf("Case#%d:",t);{}for(N=4;N<=10000;N++)//4到10000直 進}return}HDU49522014xkkxxk的倍x。(1<=xk<=10^10)#include#include<cstdio>longlongx,s;int;intmain(){while(1){if(x==0&&s==0)break;for(inti=1;i<s;i++){if(x<i+1)}printf("Case#%d:%I64d\n",++}}SPOJ10232問在自然數(shù)中，第n3intcntfor(inti=2;i<=n;{ifni0)//{while(n%i==0{n/=}}}returncnt>=#include<iostream>#include<cstdio>usingnamespacestd;intans[1111]={…};intmain(){intn,m;{}}livearchive59861，表示他們之中有人贏了后面某幾人。3346n(n-1)/2。usingnamespacestd;intints[51],p[51];intt,i,n,res;}}}return}c/d0<c<d<busingnamespacestd;intan[100]; (inta,int{intr=0;{}return}intfenjie(inta,intb,intan[])//{intn=0;intt;{}returnn;}int{intn,A,B,C,D;{for(int{intd= {}intlen=fenjie(A,B,an);for(intj=len-2;j>=0;j--{intt=an[j]*D+C;}}}return}擴展算法是用來在已知a,b求解一組x，y使得ax+by= (a,b)=d(解一定存在，intex(inta,intb,int&x,int&y){if(b==0){x=1;y=0;return}intr=ex(b,a%b,x,intt=x;x=y;y=t-a/b*y;returnr;}1（0的話表示可以約分，直接輸出約分后的分數(shù)。分母盡可能大。a*d-b*c=1b*c-由擴展a*x+b*y=(a,b);所以d=x,c=-y或者c=y,d=-int64int64int64int64 int64 (int64a,int64b,int64&x,int64{int64if(b==0){x=1;y=0;returna;else{ans=extended_ (b,a%b,x,y);t=x;x=y;y=t-(a/b)*y;}returnans;}int{{ d1=(x+b)%b;c1=(-c2=(y+a)%a;d2=(-if(d1<d2)printf("%I64d/%I64d\n",c2,d2);elseprintf("%I64d/%I64d\n",c1,d1);}return}CodeforcesRound#266a*b6個直接，從6*n開始找出這個數(shù)能拆成的兩個數(shù)的積的可能性，然后和原先的ab進行比#include<iostream>usingnamespacestd;typedeflonglongll;intmain(){lln,a,b;lli,j;while(cin>>n>>a>>{n*=if(a*b>={cout<<a*b<<cout<<a<<""<<b<<endl;}while{for(i=min(a,b);i*i<=n;i++)//拆當前{if(n%i=={j=n/if(i>=min(a,b)&&j>=max(a,{cout<<n<<if(a>b)cout<<j<<""<<i<<endl;elsecout<<i<<""<<j<<endl;gotoend;}}}}}return}Dnk(xix平均)^2xi^2x平均^22*xi*x平均所以方差=∑xi^2+n*x平均-2*∑xi*x平均事先計算∑xi^2和∑xi，便可以以o(n)#define#define_CRT_SECUR