HDU 3404&poj 3533 Nim product (two-dimensional & three-dimensional)

(Nim product related data from the paper Cao Chinxiang "k times Dynamic subtraction game" to explore a class of combinatorial game problems)

Two function flows for the NIM product calculation:

The code is implemented as follows:

intm[2][2]={0,0,0,1};intNim_multi_power (intXinty) {    if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,s=y/m,t=y%m; intd1=Nim_multi_power (p,s); intD2=Nim_multi_power (p,t); return(m* (D1^D2)) ^nim_multi_power (m/2, D1);}intNim_multi (intXinty) {    if(x<y)returnNim_multi (y,x); if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,q=x%m,s=y/m,t=y%m; intc1=Nim_multi (p,s); intC2=nim_multi (p,t) ^Nim_multi (q,s); intc3=Nim_multi (q,t); return(m* (C1^C2)) ^c3^nim_multi_power (m/2, C1);}

Here are two game problems that are implemented with the NIM product:

(1) HDU 3404 Switch lights:http://acm.hdu.edu.cn/showproblem.php?pid=3404

#include <iostream>#include<cstdio>using namespacestd;intm[2][2]={0,0,0,1};intNim_multi_power (intXinty) {    if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,s=y/m,t=y%m; intd1=Nim_multi_power (p,s); intD2=Nim_multi_power (p,t); return(m* (D1^D2)) ^nim_multi_power (m/2, D1);}intNim_multi (intXinty) {    if(x<y)returnNim_multi (y,x); if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,q=x%m,s=y/m,t=y%m; intc1=Nim_multi (p,s); intC2=nim_multi (p,t) ^Nim_multi (q,s); intc3=Nim_multi (q,t); return(m* (C1^C2)) ^c3^nim_multi_power (m/2, C1);}intMain () {intT,n,x,y; scanf ("%d",&t);  while(t--) {scanf ("%d",&N); intres=0;  while(n--) {scanf ("%d%d",&x,&y); Res^=Nim_multi (x, y); }        if(res) printf ("There is a try, lxhgww.\n"); Elseprintf"Don ' t waste your time.\n"); }    return 0;}

(2) POJ 3533 Light switching game:http://poj.org/problem?id=3533

The previous question is a two-dimensional NIM product, this problem is different from the three-dimensional NIM product, but the same nature.

#include <iostream>#include<cstdio>using namespacestd;intm[2][2]={0,0,0,1};intNim_multi_power (intXinty) {    if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,s=y/m,t=y%m; intd1=Nim_multi_power (p,s); intD2=Nim_multi_power (p,t); return(m* (D1^D2)) ^nim_multi_power (m/2, D1);}intNim_multi (intXinty) {    if(x<y)returnNim_multi (y,x); if(x<2)        returnM[x][y]; intA=0;  for(;; a++)        if(X>= (1<< (1<<a)) &&x< (1<< (1<< (A +1))))             Break; intm=1<< (1<<a); intp=x/m,q=x%m,s=y/m,t=y%m; intc1=Nim_multi (p,s); intC2=nim_multi (p,t) ^Nim_multi (q,s); intc3=Nim_multi (q,t); return(m* (C1^C2)) ^c3^nim_multi_power (m/2, C1);}intNim_multi2 (intXintYintz) {    intt=Nim_multi (x, y); returnNim_multi (t,z);}intMain () {intn,x,y,z;  while(~SCANF ("%d",&N)) {intres=0;  while(n--) {scanf ("%d%d%d",&x,&y,&z); Res^=nim_multi2 (x, y, z); }        if(res) printf ("no\n"); Elseprintf"yes\n"); }    return 0;}

HDU 3404&poj 3533 Nim product (two-dimensional & three-dimensional)