ZOJ3380 Patchouli's Spell Cards C++版(概率DP+大数)

题意题解不说了，这里主要讲用C++大数

为了节省时间使用了进制压缩，使用long long数组，进制1e9，使进制的平方在long long范围内

用len记数字位数，保证无前导零，方便计算，最低位存在number[0]中，number数组开到26(开25返回Segmentation Fault)

operator定义运算规则，这里+ - * / % == !=都进行了定义

1290ms 4836KB

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const long long mod=1000000000;
struct biginteger
{long long number[26];int len;biginteger operator+(biginteger &b){int i,root=max(len,b.len),Len=1;biginteger ans;memset(ans.number,0,sizeof(ans.number));ans.len=1;for(i=0; i<root||ans.number[i]; i++){ans.number[i]+=number[i]+b.number[i];if(ans.number[i]>=mod){ans.number[i+1]+=ans.number[i]/mod;ans.number[i]%=mod;}if(ans.number[i])Len=i;}ans.len=Len+1;return ans;}biginteger operator-(biginteger &b){int i,root=max(len,b.len),Len=0;biginteger ans;memset(ans.number,0,sizeof(ans.number));ans.len=1;for(i=0; i<root; i++){ans.number[i]+=number[i]-b.number[i];if(ans.number[i]<0){ans.number[i+1]-=1;ans.number[i]+=mod;}if(ans.number[i])Len=i;}ans.len=Len+1;return ans;}biginteger operator*(biginteger &b){int i,j,root=0;biginteger ans;memset(ans.number,0,sizeof(ans.number));ans.len=1;for(i=0; i<len; i++){for(j=0; j<b.len; j++)ans.number[i+j]+=number[i]*b.number[j];}for(i=0; i<=len+b.len||ans.number[i]; i++){if(ans.number[i]>=mod){ans.number[i+1]+=ans.number[i]/mod;ans.number[i]%=mod;}if(ans.number[i])root=i;}ans.len=root+1;return ans;}bool Greater(biginteger b){if(len>b.len)return true;if(len<b.len)return false;int i;for(i=len-1; i>=0; i--){if(number[i]>b.number[i])return true;if(number[i]<b.number[i])return false;}return true;}biginteger operator/(biginteger &b){biginteger ans_quotient;biginteger ans_remainder;memset(ans_quotient.number,0,sizeof(ans_quotient.number));ans_quotient.len=1;memset(ans_remainder.number,0,sizeof(ans_remainder.number));ans_remainder.len=1;biginteger temp;int i,j,root=0;long long w,head=0,l,r,mid;for(i=len-1; i>=b.len-1; i--){head*=mod;head+=number[i];l=head/(b.number[b.len-1]+1);r=head/b.number[b.len-1];while(l<=r){mid=(l+r)/2;memset(temp.number,0,sizeof(temp.number));temp.len=1;temp.number[i-b.len+1]=mid;temp.len=i-b.len+2;temp=temp*b;if(Greater(temp))l=mid+1;elser=mid-1;}w=r;memset(temp.number,0,sizeof(temp.number));temp.len=1;temp.number[i-b.len+1]=w;temp.len=i-b.len+2;temp=temp*b;int xroot=max(len,temp.len);for(j=0; j<xroot; j++){number[j]-=temp.number[j];if(number[j]<0){number[j+1]-=1;number[j]+=mod;}if(number[j])len=j+1;}ans_quotient.number[i-b.len+1]=w;head=number[i];if(root==0&&w!=0)root=i-b.len+2;}ans_quotient.len=max(root,1);for(i=0; i<len; i++)ans_remainder.number[i]=number[i];ans_remainder.len=1;for(i=len-1; i>=0; i--){if(ans_remainder.number[i]){ans_remainder.len=i+1;break;}}return ans_quotient;}biginteger operator%(biginteger &b){biginteger ans_quotient;biginteger ans_remainder;memset(ans_quotient.number,0,sizeof(ans_quotient.number));ans_quotient.len=1;memset(ans_remainder.number,0,sizeof(ans_remainder.number));ans_remainder.len=1;biginteger temp;int i,j,root=0;long long w,head=0,l,r,mid;for(i=len-1; i>=b.len-1; i--){head*=mod;head+=number[i];l=head/(b.number[b.len-1]+1);r=head/b.number[b.len-1];while(l<=r){mid=(l+r)/2;memset(temp.number,0,sizeof(temp.number));temp.len=1;temp.number[i-b.len+1]=mid;temp.len=i-b.len+2;temp=temp*b;if(Greater(temp))l=mid+1;elser=mid-1;}w=r;memset(temp.number,0,sizeof(temp.number));temp.len=1;temp.number[i-b.len+1]=w;temp.len=i-b.len+2;temp=temp*b;int xroot=max(len,temp.len);for(j=0; j<xroot; j++){number[j]-=temp.number[j];if(number[j]<0){number[j+1]-=1;number[j]+=mod;}if(number[j])len=j+1;}ans_quotient.number[i-b.len+1]=w;head=number[i];if(root==0&&w!=0)root=i-b.len+2;}ans_quotient.len=max(root,1);for(i=0; i<len; i++)ans_remainder.number[i]=number[i];ans_remainder.len=1;for(i=len-1; i>=0; i--){if(ans_remainder.number[i]){ans_remainder.len=i+1;break;}}return ans_remainder;}bool operator==(biginteger &b){if(len!=b.len)return false;int i;for(i=0; i<len; i++){if(number[i]!=b.number[i])return false;}return true;}bool operator!=(biginteger &b){if(len!=b.len)return true;int i;for(i=0; i<len; i++){if(number[i]!=b.number[i])return true;}return false;}
} c[105][105],dp[105][105],zero,one;
void Print(biginteger a);
biginteger Gcd(biginteger a,biginteger b)
{while(b!=zero){biginteger temp=b;b=a%b;a=temp;}return a;
}
void Print(biginteger a)
{int i;printf("%lld",a.number[a.len-1]);for(i=a.len-2; i>=0; i--)printf("%09lld",a.number[i]);
}
int main()
{one.len=1;one.number[0]=1;zero.len=1;int i,j,k;for(i=0; i<=100; i++)c[i][0]=c[i][i]=one;for(i=2; i<=100; i++){for(j=1; j<i; j++)c[i][j]=c[i-1][j-1]+c[i-1][j];}long long n,m,l;while(~scanf("%lld%lld%lld",&m,&n,&l)){if(l>m){printf("mukyu~\n");continue;}for(i=0; i<=100; i++){for(j=0; j<=100; j++)dp[i][j]=zero;}dp[0][0]=one;biginteger ans=zero;for(i=1; i<=n; i++){for(j=1; j<=m; j++){for(k=0; k<=j&&k<l; k++){biginteger temp=dp[i-1][j-k]*c[m-(j-k)][k];dp[i][j]=dp[i][j]+temp;}}ans=ans+dp[i][m];}biginteger sum=one;for(i=0; i<m; i++){biginteger x=zero;x.number[0]=n;sum=sum*x;}ans=sum-ans;biginteger gcd=Gcd(sum,ans);Print(ans/gcd);printf("/");Print(sum/gcd);printf("\n");}return 0;
}

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