Bzoj 2186 [Sdoi2008] salad Princess puzzle "inverse meta"

Test Instructions: the number of coprime in the request, wherein.

Analysis: because, so, we are easy to know as to the conclusion

for two positive integers and, if they are multiples, then the number of

This conclusion is well proven, because the number of the in and of the intertextuality is, and know, so

conclusion is established. So for the subject, the answer is

In fact, as long as the inverse of the prime number with EXGCD to beg for a good, the rest is not used

Inverse element Recursive method:

#include <stdio.h>#include<string.h>Const intn=1e7+ the; typedefLong Longll;intPr[n],p[n],cnt,mod;intInv[n],ans1[n],ans2[n];intRead () {intx=0;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') ch=GetChar ();  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx;}voidinit () {ans1[1]=ans2[1]=inv[1]=1;  for(intI=2; i<n;i++) {Ans1[i]= (LL) ans1[i-1]*i%MoD; if(!P[i]) pr[++cnt]=i;  for(intj=1; j<=cnt&&i*pr[j]<n;j++) {P[pr[j]*i]=1; if(i%pr[j]==0) Break; }    }     for(intI=2; i<n&&i<mod;i++) {Inv[i]= (mod-(ll) mod/i) *inv[mod%i]%MoD; }     for(intI=2; i<n;i++) {Ans2[i]=ans2[i-1]; if(!P[i]) ans2[i]= (ll) ans2[i]* (i-1)%mod*inv[i%mod]%MoD; }}intMain () {intt,n,m; scanf ("%d%d",&t,&MoD);    Init ();  while(t--) {n=read (); m=read (); printf ("%d\n", (LL) ans1[n]*ans2[m]%MoD); }    return 0;}

Expand Euclid to seek inverse yuan

#include <stdio.h>#include<string.h>Const intn=1e7+ the; typedefLong Longll;intPr[n],p[n],cnt,mod;intInv[n],ans1[n],ans2[n];intRead () {intx=0;CharCh=GetChar ();  while(ch<'0'|| Ch>'9') ch=GetChar ();  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx;}intEX_GCD (intAintBint&x,int&y) {    if(!b) {x=1, y=0; returnA; }    intANS=EX_GCD (b,a%b,y,x); Y-=a/b*x; returnans;}intGETINV (inti) {    intx, y;    EX_GCD (I,mod,x,y); X= ((x%mod) +mod)%MoD; returnx;}voidinit () {ans1[1]=ans2[1]=inv[1]=1;  for(intI=2; i<n;i++) {Ans1[i]= (LL) ans1[i-1]*i%MoD; if(!P[i]) pr[++cnt]=i,inv[i]=GETINV (i);  for(intj=1; j<=cnt&&i*pr[j]<n;j++) {P[pr[j]*i]=1; if(i%pr[j]==0) Break; }    }     for(intI=2; i<n;i++) {Ans2[i]=ans2[i-1]; if(!P[i]) ans2[i]= (ll) ans2[i]* (i-1)%mod*inv[i%mod]%MoD; }}intMain () {intt,n,m; scanf ("%d%d",&t,&MoD);    Init ();  while(t--) {n=read (); m=read (); printf ("%d\n", (LL) ans1[n]*ans2[m]%MoD); }    return 0;}

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http://blog.csdn.net/acdreamers/article/details/8220787

Bzoj 2186 [Sdoi2008] salad Princess puzzle "inverse meta"