bzoj 3151: [Ctsc2013]因式分解 数学

       假设分解为∏(pix+qi)，那么就有∏pi=An，∏qi=A0。由于有pi<=1000000，qi<=1000000，我们枚举An和A0的约数然后暴力枚举判断。。。

       至于于是判断，可以考虑在模意义下判断，选70个大质数分别判断即可。。。

       剩下的就是码码码了。。

AC代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define ll long longusing namespace std;const int prm[70]={543892411,567498259,643448581,772660877,802279559,823153213,847485637,907170331,919507003,923425961,929812711,935151689,938910911,962946403,970870897,977946113,985116131,989743967,991308817,994793521,995403061,997310857,998623607,999049111,999860857,1000885111,1002002609,1002270383,1002326077,1002351107,1002642457,1002681557,1003013519,1004475841,1005879367,1006478441,1007120461,1007468677,1007576071,1008054149,1008640013,1008894637,1009316083,1009528721,1009862393,1010510329,1010915173,1011460739,1012072651,1012612079,1012668131,1012825181,1014091867,1014515053,1014904973,1015114333,1015944323,1016003431,1016121461,1017088367,1017307463,1017423811,1017620873,1018394743,1018932493,1019199101,1019208581,1019319187,1019376101,1019831203};int n,m,cnt,tot1,tot2,b[70][85],p[100005],q[100005];char s0[30005],s1[30005];struct node{ int x,y,z; }ans[85];bool cmp(const node &u,const node &v){return (ll)u.x*v.y>(ll)u.y*v.x;}struct hugnum{int len,sgn,num[40];void ins(int tot,int fu){int i,j; sgn=fu;for (i=tot; i>0; i-=9){len++;for (j=max(1,i-8); j<=i; j++) num[len]=num[len]*10+s1[j]-'0';}}int getmod(int x){int i,t=0;for (i=len; i; i--)t=((ll)t*1000000000+num[i])%x;return t;}}a[85];void get_in(){scanf("%s",s0+1);int i=1,len=strlen(s0+1);while (i<=len){int tot=0,fu=1,dgr=0;if (s0[i]=='-') fu=-1;if (s0[i]=='+' || s0[i]=='-') i++;if (s0[i]=='x'){s1[tot=1]='1'; i++;if (s0[i]=='^'){for (i++; s0[i]>='0' && s0[i]<='9'; i++)dgr=dgr*10+s0[i]-'0';a[dgr].ins(tot,fu);if (!n) n=dgr;} else a[1].ins(tot,fu);} else{for (; s0[i]>='0' && s0[i]<='9'; i++) s1[++tot]=s0[i];if (!s0[i]) a[0].ins(tot,fu); else{i++;if (s0[i]=='^'){for (i++; s0[i]>='0' && s0[i]<='9'; i++)dgr=dgr*10+s0[i]-'0';a[dgr].ins(tot,fu);if (!n) n=dgr;} else a[1].ins(tot,fu);}}}if (a[n].sgn<0) putchar('-');if (a[n].len>1 || a[n].num[1]>1){printf("%d",a[n].num[a[n].len]);for (i=a[n].len-1; i; i--) printf("%09d",a[n].num[i]);}}void put_out(){sort(ans+1,ans+cnt+1,cmp); int i;for (i=1; i<=cnt; i++){if (!ans[i].x) putchar('x'); elseif (ans[i].y==1)if (ans[i].x<0) printf("(x+%d)",-ans[i].x);else printf("(x%d)",-ans[i].x);elseif (ans[i].x<0) printf("(x+%d/%d)",-ans[i].x,ans[i].y);else printf("(x%d/%d)",-ans[i].x,ans[i].y);if (ans[i].z>1) printf("^%d",ans[i].z);}puts("");}int ksm(int x,int y,int mod){int t=1; for (; y; y>>=1,x=(ll)x*x%mod) if (y&1) t=(ll)t*x%mod;return t;}int gcd(int x,int y){ return (y)?gcd(y,x%y):x; }bool ok(int x,int y){int i,j,t,mod,tmp;for (i=0; i<70; i++){mod=prm[i]; t=(x+mod)%mod;t=(ll)t*ksm(y,mod-2,mod)%mod;for (j=n,tmp=0; j>=0; j--)tmp=((ll)tmp*t+b[i][j])%mod;if (tmp) return 0;}return 1;}void divide(int x,int y){int i,j,t,mod;for (i=0; i<70; i++){mod=prm[i]; t=(x+mod)%mod;t=(ll)t*ksm(y,mod-2,mod)%mod;for (j=n; j>=0; j--)b[i][j]=((ll)b[i][j+1]*t+b[i][j])%mod;}}int main(){get_in();int i=0,j,k; while (!a[i].len) i++;if (i>0) ans[++cnt]=(node){0,1,i};for (j=i; j<=n; j++) a[j-i]=a[j];n-=i;for (i=0; i<70; i++)for (j=0; j<=n; j++) b[i][j]=(a[j].getmod(prm[i])*a[j].sgn+prm[i])%prm[i];for (i=1; i<=1000000; i++){if (!a[0].getmod(i)) p[++tot1]=i;if (!a[n].getmod(i)) q[++tot2]=i;}for (i=1; i<=tot1; i++)for (j=1; j<=tot2; j++) if (gcd(p[i],q[j])==1){for (k=0; ok(p[i],q[j]); k++)divide(p[i],q[j]);if (k) ans[++cnt]=(node){p[i],q[j],k};for (k=0; ok(-p[i],q[j]); k++)divide(-p[i],q[j]);if (k) ans[++cnt]=(node){-p[i],q[j],k};}put_out();return 0;}

by lych

2016.5.26