gcd，lcm，ext_gcd，inv
Least Common Multiple http://acm.hdu.edu.cn/showproblem.php?pid=1019
1 #include<cstdio> 2 int gcd(int a,int b){ 3 return b?gcd(b,a%b):a; 4 } 5 int lcm(int a,int b){ 6 return a/gcd(a,b)*b; 7 } 8 int main(){ 9 int n,m,ans,x; 10 while(~scanf("%d",&n)){ 11 while(n--){ 12 ans=1; 13 scanf("%d",&m); 14 while(m--){ 15 scanf("%d",&x); 16 ans=lcm(ans,x); 17 } 18 printf("%d\n",ans); 19 } 20 } 21 return 0; 22 }
Turn the pokers http://acm.hdu.edu.cn/showproblem.php?pid=4869
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 typedef __int64 LL; 5 const int M=100010; 6 const int mod=1000000009; 7 int ext_gcd(int a,int b,int &x,int &y){//扩展gcd d=gcd(a,b)=a*x+b*y; return d,x,y; 8 int t,ret; 9 if(!b){ 10 x=1; 11 y=0; 12 return a; 13 } 14 ret=ext_gcd(b,a%b,x,y); 15 t=x; 16 x=y; 17 y=t-a/b*y; 18 return ret; 19 } 20 int inv(int a,int b,int c){//ext_gcd求逆元 (b/a)%c 21 int x,y; 22 ext_gcd(a,c,x,y); 23 return (1LL*x*b%c+c)%c; 24 } 25 LL C[M]; 26 LL INV[M]; 27 int main() { 28 for(int i=1; i<M; i++) { 29 INV[i]=inv(i,1,mod); 30 } 31 int n,m,a; 32 while(~scanf("%d%d",&n,&m)) { 33 C[0]=1; 34 for(int i=1;i<=m;i++){ 35 C[i]=C[i-1]*(m-i+1)%mod*INV[i]%mod; 36 } 37 int L=0,R=0,nl,nr,tmp; 38 for(int i=0;i<n;i++){ 39 scanf("%d",&a); 40 tmp=min(m-L,a); 41 nr=L+tmp-(a-tmp); 42 tmp=min(R,a); 43 nl=R-tmp+(a-tmp); 44 if(nl>nr) swap(nl,nr); 45 if(L<=a&&a<=R){ 46 if(L%2==a%2){ 47 nl=0; 48 } 49 else{ 50 nl=min(nl,1); 51 } 52 } 53 if((m-R)<=a&&a<=(m-L)){ 54 if((m-L)%2==a%2){ 55 nr=m; 56 } 57 else{ 58 nr=max(nr,m-1); 59 } 60 } 61 if(L>=a) nl=min(nl,L-a); 62 if(m-R>=a) nr=max(nr,R+a); 63 L=nl; 64 R=nr; 65 } 66 int ans=0; 67 for(int i=L;i<=R;i+=2){ 68 ans+=C[i]; 69 ans%=mod; 70 } 71 printf("%d\n",ans); 72 } 73 return 0; 74 }
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 typedef __int64 LL; 5 const int M=100010; 6 const int mod=1000000009; 7 LL C[M]; 8 LL INV[M]; 9 void inv_init(){//初始化%mod的乘法逆元 10 INV[1]=1; 11 for(int i=2;i<M;i++){ 12 INV[i]=INV[mod%i]*(mod-mod/i)%mod; 13 } 14 } 15 int main() { 16 inv_init(); 17 int n,m,a; 18 while(~scanf("%d%d",&n,&m)) { 19 C[0]=1; 20 for(int i=1;i<=m;i++){ 21 C[i]=C[i-1]*(m-i+1)%mod*INV[i]%mod; 22 } 23 int L=0,R=0,nl,nr,tmp; 24 for(int i=0;i<n;i++){ 25 scanf("%d",&a); 26 tmp=min(m-L,a); 27 nr=L+tmp-(a-tmp); 28 tmp=min(R,a); 29 nl=R-tmp+(a-tmp); 30 if(nl>nr) swap(nl,nr); 31 if(L<=a&&a<=R){ 32 if(L%2==a%2){ 33 nl=0; 34 } 35 else{ 36 nl=min(nl,1); 37 } 38 } 39 if((m-R)<=a&&a<=(m-L)){ 40 if((m-L)%2==a%2){ 41 nr=m; 42 } 43 else{ 44 nr=max(nr,m-1); 45 } 46 } 47 if(L>=a) nl=min(nl,L-a); 48 if(m-R>=a) nr=max(nr,R+a); 49 L=nl; 50 R=nr; 51 } 52 int ans=0; 53 for(int i=L;i<=R;i+=2){ 54 ans+=C[i]; 55 ans%=mod; 56 } 57 printf("%d\n",ans); 58 } 59 return 0; 60 }
end
