UVa:11375 Matches

来源：互联网发布：mac 16进制编辑器编辑：程序博客网时间：2024/06/07 09:07

白书上的例题。用到了大数加法。注意它问的是用n根火柴可以摆成多少个整数，就是说少于n根火柴的时候也是可以的。另外注意前导0的情况。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <vector>#include <queue>#include <algorithm>#define ll long long#define INF 2139062143#define MOD 20071027#define MAXN 20005using namespace std;int v[]= {6,2,5,5,4,5,6,3,7,6};int w[]= {0,0,1,1,1,2,3,1,0,0};void BigAdd(char *a,char *b,char *back){    char c[500]= {0};    int p=0,up=0;    for(int i=strlen(a)-1,j=strlen(b)-1; i>=0||j>=0; --i,--j)    {        int x,y,z;        if(i<0) x=0;        else x=a[i]-'0';        if(j<0) y=0;        else y=b[j]-'0';        z=(x+y+up)%10;        up=(x+y+up)/10;        c[p++]=z+'0';    }    if(up) c[p++]='1';    c[p]=0;    int q=0;    for(int i=p-1; i>=0; --i)        back[q++]=c[i];    if(q==0) back[q++]='0';    back[q]=0;}char f[2010][500];char d[2010][500];char temp[500]= {0};void Init(){    strcpy(d[0],"1");    for(int i=1; i<=2000; ++i)strcpy(d[i],"0");    for(int i=0; i<=2000; ++i)        for(int j=0; j<10; ++j)            if(!(i==0&&j==0))            BigAdd(d[i+v[j]],d[i],d[i+v[j]]);    BigAdd(d[6],"1",d[6]);    strcpy(f[0],"0");    for(int i=1; i<=2000; ++i)        BigAdd(d[i],f[i-1],f[i]);}int main(){    Init();    int n;    while(scanf("%d",&n)!=EOF)        puts(f[n]);    return 0;}

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