HDOJ 5794 (2016多校联合训练 Training Contest 6) A Simple Chess
2016-08-06 14:45
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5794
第六场多校唯一"做"出来的一题(小编觉得,像 A Boring Question(01)和 A Simple Nim(03)这种打个表莫名其妙就A的题没啥好讲的)。
题意:给我们一个棋盘,棋盘上有r个障碍,我们能够从(1,1)出发,每一次能够向右下走日字,问我们走到终点的方案数。
很简单的一个容斥原理的运用,举个例子,我们从点i出发,途中有一个障碍点j,从点i到终点的方案数,就等于总方案数减去从j到终点的方案数*从点i到点j的方案数。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL mod=110119;
const LL maxn=100;
LL dp[maxn*10+20];
LL f[mod*10];
struct Point{LL x,y;}p[maxn+20];
bool cmp(Point a,Point b)
{
if(a.x<b.x)return 1;
if(a.x>b.x)return 0;
return a.y<b.y;
}
LL Power_mod(LL a, LL b, LL p)
{
LL res = 1;
while(b)
{
if(b&1) res = (res * a) % p;
a = (a*a) % p;
b >>= 1;
}
return res;
}
LL Comb(LL a, LL b, LL p)
{
if(a < b) return 0;
if(a == b) return 1;
LL ans=f[a]*Power_mod(f[b]*f[a-b]%p,p-2,p)%p;
return ans;
}
LL Lucas(LL n, LL m, LL p)
{
LL ans = 1;
while(n && m && ans)
{
ans = (ans*Comb(n%p, m%p, p)) % p;
n /= p;
m /= p;
}
return ans;
}
LL get(LL x,LL y)
{
if((x+y)%3 || max(x,y)>2*min(x,y))return 0;
LL a=(x+y)/3,b=(2*x-y)/3;
if(a<0 || b<0)return 0;
return Lucas(a,b,mod);
}
int main()
{
LL i,j,k,h,w,n,cas=0;
f[0]=f[1]=1;
for(i=2;i<=mod+20;i++)f[i]=f[i-1]*i%mod;
while(~scanf("%I64d%I64d%I64d",&h,&w,&n))
{
for(i=0; i<n; i++) scanf("%I64d%I64d",&p[i].x,&p[i].y);
if((h+w-2)%3 || max(h-1,w-1)>2*min(h-1,w-1))
{
printf("Case #%I64d: 0\n",++cas);
continue;
}
sort(p,p+n,cmp),p
.x=h,p
.y=w;
for(i=0; i<=n; i++)
{
dp[i] = get(p[i].x-1,p[i].y-1);
for(j=0; j<i; j++)
if(p[j].x <= p[i].x && p[j].y <= p[i].y)
{
dp[i] -= get(p[i].x-p[j].x,p[i].y-p[j].y)*dp[j];
dp[i] = (dp[i]%mod+mod)%mod;
}
}
printf("Case #%I64d: %I64d\n",++cas,dp
);
}
return 0;
}
第六场多校唯一"做"出来的一题(小编觉得,像 A Boring Question(01)和 A Simple Nim(03)这种打个表莫名其妙就A的题没啥好讲的)。
题意:给我们一个棋盘,棋盘上有r个障碍,我们能够从(1,1)出发,每一次能够向右下走日字,问我们走到终点的方案数。
很简单的一个容斥原理的运用,举个例子,我们从点i出发,途中有一个障碍点j,从点i到终点的方案数,就等于总方案数减去从j到终点的方案数*从点i到点j的方案数。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL mod=110119;
const LL maxn=100;
LL dp[maxn*10+20];
LL f[mod*10];
struct Point{LL x,y;}p[maxn+20];
bool cmp(Point a,Point b)
{
if(a.x<b.x)return 1;
if(a.x>b.x)return 0;
return a.y<b.y;
}
LL Power_mod(LL a, LL b, LL p)
{
LL res = 1;
while(b)
{
if(b&1) res = (res * a) % p;
a = (a*a) % p;
b >>= 1;
}
return res;
}
LL Comb(LL a, LL b, LL p)
{
if(a < b) return 0;
if(a == b) return 1;
LL ans=f[a]*Power_mod(f[b]*f[a-b]%p,p-2,p)%p;
return ans;
}
LL Lucas(LL n, LL m, LL p)
{
LL ans = 1;
while(n && m && ans)
{
ans = (ans*Comb(n%p, m%p, p)) % p;
n /= p;
m /= p;
}
return ans;
}
LL get(LL x,LL y)
{
if((x+y)%3 || max(x,y)>2*min(x,y))return 0;
LL a=(x+y)/3,b=(2*x-y)/3;
if(a<0 || b<0)return 0;
return Lucas(a,b,mod);
}
int main()
{
LL i,j,k,h,w,n,cas=0;
f[0]=f[1]=1;
for(i=2;i<=mod+20;i++)f[i]=f[i-1]*i%mod;
while(~scanf("%I64d%I64d%I64d",&h,&w,&n))
{
for(i=0; i<n; i++) scanf("%I64d%I64d",&p[i].x,&p[i].y);
if((h+w-2)%3 || max(h-1,w-1)>2*min(h-1,w-1))
{
printf("Case #%I64d: 0\n",++cas);
continue;
}
sort(p,p+n,cmp),p
.x=h,p
.y=w;
for(i=0; i<=n; i++)
{
dp[i] = get(p[i].x-1,p[i].y-1);
for(j=0; j<i; j++)
if(p[j].x <= p[i].x && p[j].y <= p[i].y)
{
dp[i] -= get(p[i].x-p[j].x,p[i].y-p[j].y)*dp[j];
dp[i] = (dp[i]%mod+mod)%mod;
}
}
printf("Case #%I64d: %I64d\n",++cas,dp
);
}
return 0;
}
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