比赛地址：https://www.nowcoder.com/acm/contest/129#question
BLOG：http://www.elijahqi.win/archives/3836
A.序列 发现0.5和-2是一对 共同出现偶数次即可抵消影响 1是凑数的
于是暴力dp dp[i][j][k]前i个位置选了j个0.5 k个-2的方案
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e3+10;
const int mod=1e9+7;
int dp[2][N][N],n;
inline int inc(int x,int v){
return x+v>=mod?x+v-mod:x+v;
}
int main(){
scanf("%d",&n);
dp[0][0][0]=1;int pre=0,now=1;
for (int i=0;i<n-1;++i){
memset(dp[now],0,sizeof(dp[now]));
for (int j=0;j<=i;++j){
for (int k=0;k<=i-j;++k){
dp[now][j][k]=inc(dp[now][j][k],dp[pre][j][k]);
dp[now][j+1][k]=inc(dp[now][j+1][k],dp[pre][j][k]);
dp[now][j][k+1]=inc(dp[now][j][k+1],dp[pre][j][k]);
}
}pre^=1;now^=1;
}int ans=0;
for (int i=0;i*4<=n-1;++i)ans=inc(ans,dp[pre][i<<1][i<<1]);printf("%d\n",ans);
return 0;
}
实际上组合数学也可以搞 就枚举一下这个偶数到底是多少然后分别钦定一下-2和0.5的位置即可
参照大爷代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 1010
#define mod 1000000007
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}
int n,C[N][N],ans=0;
inline void inc(int &x,int y){x+=y;x%=mod;}
int main(){
// freopen("a.in","r",stdin);
n=read()-1;
for(int i=0;i<=n;++i) C[i][0]=1;
for(int i=1;i<=n;++i)
for(int j=1;j<=i;++j)
C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
for(int i=0;i*4<=n;++i)
inc(ans,(ll)C[n][2*i]*C[n-2*i][2*i]%mod);
printf("%d\n",ans);
return 0;
}
B.随机数
设f[i]表示前i个数有奇数个1的概率，则
f[i+1]=f[i]∗(1−p)+(1−f[i])∗pf[i+1]=f[i]∗(1−p)+(1−f[i])∗p
f[i+1]=(1−2p)f[i]+pf[i+1]=(1−2p)f[i]+p
然后十进制矩阵快速幂搞掉即可 每次倍增十次
就从比如10^10变成10^100
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
const int N=1e6+10;
inline int ksm(ll b,int t){
static ll tmp;
for (tmp=1;t;b=b*b%mod,t>>=1) if (t&1) tmp=tmp*b%mod;return tmp;
}
int a;char s[N];
struct matrix{
int f[3][3];
}tr,ans;
inline int inc(int x,int v){return x+v>=mod?x+v-mod:x+v;}
inline matrix multiply(const matrix &a,const matrix &b){
matrix c;memset(c.f,0,sizeof(c.f));
for (int i=1;i<=2;++i){
for (int j=1;j<=2;++j){
for (int k=1;k<=2;++k){
c.f[i][j]=inc(c.f[i][j],(ll)a.f[i][k]*b.f[k][j]%mod);
}
}
}return c;
}
inline matrix ksm(matrix b,int t){
matrix res;memset(res.f,0,sizeof(res.f));res.f[1][1]=res.f[2][2]=1;
for (;t;b=multiply(b,b),t>>=1) if (t&1) res=multiply(res,b);return res;
}
int main(){
freopen("b.in","r",stdin);
scanf("%d",&a);
scanf("%s",s+1);
int inva=1ll*a*ksm(10000,mod-2)%mod;
int invb=1-inva+mod;invb%=mod;
tr.f[1][1]=invb;tr.f[1][2]=inva;
tr.f[2][1]=inva;tr.f[2][2]=invb;
int n=strlen(s+1);ans.f[1][1]=ans.f[2][2]=1;
for (int i=n;i;--i){
int p=s[i]-'0';matrix tmp=tr;
tmp=ksm(tmp,p);
tr=ksm(tr,10);ans=multiply(ans,tmp);
}printf("%d\n",ans.f[1][2]);
return 0;
}
c求全局到某个点距离即可 曼哈顿距离其实可以针对每个维度分开维护
菜鸡elijahqi代码写复杂了
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2200;
const int mod=1e9+7;
char s[N][N];
int tot,n,m;
int prex[N],prey[N],sufx[N],sufy[N];
int nprex[N],nprey[N],nsufx[N],nsufy[N],inv;
struct node{
int x,y;
}pt[N*N];
inline int ksm(ll b,int t){
static ll tmp;
for (tmp=1;t;b=b*b%mod,t>>=1) if (t&1) tmp=tmp*b%mod;return tmp;
}
inline bool cmpx(const node &a,const node &b){
return a.x<b.x;
}
inline bool cmpy(const node &a,const node &b){
return a.y<b.y;
}
inline int inc(int x,int v){
return x+v>=mod?x+v-mod:x+v;
}
inline int dec(int x,int v){
return x-v<0?x-v+mod:x-v;
}
inline int calc(int x,int y){
int sum=0;
for (int i=1;i<=n;++i){
for (int j=1;j<=m;++j) if(s[i][j]=='1') sum=inc(sum,abs(x-i)+abs(y-j));
}return sum;
}
inline int gao(int x,int y){
int sum=0;sum=inc(sum,dec(1ll*x*nprex[x]%mod,prex[x]));
sum=inc(sum,dec(1ll*(n-x)*nsufx[x]%mod,sufx[x]));
sum=inc(sum,dec(1ll*y*nprey[y]%mod,prey[y]));
sum=inc(sum,dec(1ll*(m-y)*nsufy[y]%mod,sufy[y]));
//printf("%d\n",sum);printf("%d\n",calc(x,y));
//sum=calc(x,y);
sum=(ll)sum*inv%mod;return sum;
}
int main(){
freopen("c.in","r",stdin);
scanf("%d%d",&n,&m);
for (int i=1;i<=n;++i) scanf("%s",s[i]+1);
for (int i=1;i<=n;++i)
for (int j=1;j<=m;++j)
if (s[i][j]=='1') pt[++tot].x=i,pt[tot].y=j;
sort(pt+1,pt+tot+1,cmpx);int now=1;inv=ksm(tot,mod-2);
for (int i=1;i<=n;++i){
prex[i]=prex[i-1];nprex[i]=nprex[i-1];
while(now<=tot&&pt[now].x<=i){
prex[i]+=pt[now].x;prex[i]%=mod;
++nprex[i];++now;
}
}now=tot;//printf("%lld\n",1ll*16*ksm(7,mod-2)%mod);
for (int i=n;i;--i){
sufx[i]=sufx[i+1];nsufx[i]=nsufx[i+1];
while(now&&pt[now].x>=i){
sufx[i]+=n-pt[now].x;sufx[i]%=mod;
++nsufx[i];--now;
}
}sort(pt+1,pt+tot+1,cmpy);
now=1;
for (int i=1;i<=m;++i){
prey[i]=prey[i-1];nprey[i]=nprey[i-1];
while(now<=tot&&pt[now].y<=i){
prey[i]+=pt[now].y;prey[i]%=mod;
++nprey[i];++now;
}
}now=tot;
for (int i=m;i;--i){
sufy[i]=sufy[i+1];nsufy[i]=nsufy[i+1];
while(now&&pt[now].y>=i){
sufy[i]+=m-pt[now].y;sufy[i]%=mod;
++nsufy[i];--now;
}
}int ans=0;
for (int i=1;i<=n;++i){
for (int j=1;j<=m;++j){
//printf("%d\n",gao(i,j));
ans^=gao(i,j);
}
}printf("%d\n",ans);
return 0;
}
icefox大爷的代码(icefox天下第一！
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 2010
#define mod 1000000007
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x*f;
}
int n,m,s1[N],s2[N],tot,f1[N],f2[N],ans=0;
char s[N];
inline void inc(int &x,int y){x+=y;x%=mod;}
inline int ksm(int x,int k){
int res=1;for(;k;k>>=1,x=(ll)x*x%mod) if(k&1) res=(ll)res*x%mod;return res;
}
int main(){
// freopen("a.in","r",stdin);
n=read();m=read();
for(int i=1;i<=n;++i){
scanf("%s",s+1);
for(int j=1;j<=m;++j){
if(s[j]=='0') continue;
s1[i]++;s2[j]++;++tot;
}
}for(int i=2;i<=n;++i) inc(f1[1],(i-1)*s1[i]);
int tmp=s1[1];
for(int i=2;i<=n;++i){
f1[i]=f1[i-1]-(tot-tmp)+tmp;f1[i]=(f1[i]%mod+mod)%mod;
tmp+=s1[i];
}for(int i=2;i<=m;++i) inc(f2[1],(i-1)*s2[i]);
tmp=s2[1];
for(int i=2;i<=m;++i){
f2[i]=f2[i-1]-(tot-tmp)+tmp;f2[i]=(f2[i]%mod+mod)%mod;
tmp+=s2[i];
}tot=ksm(tot,mod-2);
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j){
ans^=(ll)(f1[i]+f2[j])*tot%mod;
}printf("%d\n",ans);
return 0;
}
d题
设dp[t][i][j]表示t时间在i,j的方案且上一次没有移动
dp1[t][i][j][k]表示ti时间在i,j的方案且上一次移动方向i是k
#include<bits/stdc++.h>
using namespace std;
int dp[2][202][202],dp1[2][202][202][4];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int n,m,T,X1,Y1,D,x2,y2;
const int mod=1e9+7;
inline int inc(int x,int v){
return x+v>=mod?x+v-mod:x+v;
}
inline int gao(int x,int y){
return max(abs(x-X1),abs(y-Y1));
}
int main(){
freopen("d.in","r",stdin);
scanf("%d%d%d%d%d%d%d%d",&n,&m,&T,&X1,&Y1,&D,&x2,&y2);
//X1=123321,Y1=123321;
dp[0][1][1]=1;int pre=0,now=1;
for (int t=0;t<T;++t){
memset(dp[now],0,sizeof(dp[now]));memset(dp1[now],0,sizeof(dp1[now]));
for (int i=1;i<=n;++i){
for (int j=1;j<=m;++j){
for (int k=0;k<4;++k){
int x=i+dx[k],y=j+dy[k];
if(x<1||x>n||y<1||y>m) continue;
dp1[now][x][y][k]=inc(dp1[now][x][y][k],dp[pre][i][j]);
}dp[now][i][j]=inc(dp[now][i][j],dp[pre][i][j]);
if (gao(i,j)>D){
int sum=0;
for (int k=0;k<4;++k){
sum=inc(sum,dp1[pre][i][j][k]);
}
for (int k=0;k<4;++k){
int x=i+dx[k],y=j+dy[k];
if(x<1||x>n||y<1||y>m) continue;
dp1[now][x][y][k]=inc(dp1[now][x][y][k],sum);
}dp[now][i][j]=inc(dp[now][i][j],sum);
}else{
//dp[now][i][j]=inc(dp[now][i][j],dp[pre][i][j]);
for (int k=0;k<4;++k){
int x=i+dx[k],y=j+dy[k];
dp[now][i][j]=inc(dp[now][i][j],dp1[pre][i][j][k]);
if(x<1||x>n||y<1||y>m) continue;
dp1[now][x][y][k]=inc(dp1[now][x][y][k],dp1[pre][i][j][k]);
}
}
}
}pre^=1;now^=1;
}int ans=0;
for (int i=0;i<4;++i) ans=inc(ans,dp1[pre][x2][y2][i]);
ans=inc(ans,dp[pre][x2][y2]);printf("%d\n",ans);
return 0;
}
