大数运算,地址: http://acm.pku.edu.cn/JudgeOnline/problem?id=2084
#include<stdio.h> #include<stdlib.h> #include<string.h> const int OneNode = 1000000; //一位里不能超过OneNode const int NodeLen = 6; //一位储存NodeLen位,和OneNode必须同时更改,输出部分格式必须跟随这里!!! const int NumMax = 500; //储存位数限制,真实位数为NumMax*6 struct BigNum { unsigned num[NumMax] ;//高位 对 下标大位 unsigned numlen ; void set(unsigned sm=0){ num[0] = sm ; numlen = 1; }//sm<OneNode void set(char *string , int strlen) { numlen = (strlen-1) / NodeLen + 1 ; memset (num , 0 , sizeof(unsigned)*numlen ); int temp , i ; for( i=strlen-1 ; i>=0 ; i-- ) { temp = i / NodeLen ; num[temp] = num[temp]*10 + string[strlen-1-i]-'0' ; } } void print() { printf("%d",num[numlen-1]); int i = numlen-1; while( i ) { i--; printf("%06d",num[i]); } printf("\n"); } };
void Add(BigNum &a,BigNum &b,BigNum &c) // a+b ->c { unsigned lenmax = a.numlen>b.numlen?a.numlen:b.numlen; c.numlen = lenmax; unsigned i,carry=0; for ( i=0 ; i<lenmax ; i++ ) { c.num[i] = carry ; if( a.numlen > i ) c.num[i]+= a.num[i]; if( b.numlen > i ) c.num[i]+= b.num[i]; carry = c.num[i] / OneNode ; c.num[i] %= OneNode ; } if ( carry ) { c.num[i] = carry ; c.numlen ++; } }
void Mul(BigNum &a,BigNum &b,BigNum &c) // a*b ->c { unsigned carry = 0 , lenmax = a.numlen+b.numlen-1 ,i,j ; unsigned __int64 temp ; c.numlen = lenmax; for ( i=0 ; i<lenmax ; i++ ) { temp = carry ; for ( j=0 ; j<a.numlen ; j++ ) { if ( i<j ) break; if ( i-j >= b.numlen ) { j = i-b.numlen ; continue; } temp += (unsigned __int64)a.num[j] * b.num[i-j] ; } carry = temp / OneNode ; c.num[i] = temp % OneNode ; } if(carry) { c.num[i] = carry ; c.numlen ++; } }
void Cpy(BigNum &a , BigNum &b) //b-->a { a.numlen=b.numlen; memcpy(a.num,b.num,sizeof(unsigned)*b.numlen); }
BigNum dp[101];
BigNum temp, mid;
void cup () {
dp[0].set ( "1", 1 ); for ( int i=1; i<101; i++ ) { dp[i].set( "0", 1 ); for ( int j=2; j<=i*2; j+=2 ) { Mul ( dp[ (j-2)/2 ], dp[ (2*i-j)/2], temp ); Add ( dp[i], temp, mid ); Cpy ( dp[i], mid ); } } }
int main () {
int n;
cup (); while ( scanf ( "%d", &n ) != EOF && n!=-1 ) { dp[n].print (); } return 0; }
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