FZU 2102Solve equation
来源:互联网 发布:2017世乒赛丁宁知乎 编辑:程序博客网 时间:2024/05/24 16:14
http://acm.fzu.edu.cn/problem.php?pid=2102
一道看起来特别牛逼但是却很简单的题。给你一个等式A=K*B+d,给你A,B,C,C为当前进制数,让你求出满足等式的k与D.(k,d)
看似有进制转换跟着,但并不难,K为A/B(转换为10进制之后),d=A%B
之后就都懂了
。
#include <stdio.h>#include <algorithm>#include <iostream>#include <string.h>#define Max 2188#include <map>using namespace std;int convert(int scale,int x){ int sum=1; for(int i=0;i<x;i++) sum*=scale; return sum;}int account(char a[],int suma,int scale){ for(int i=strlen(a)-1,j=0;i>=0;i--,j++) { if(a[i]>='0'&&a[i]<='9') suma+=(a[i]-'0')*convert(scale,j); else { switch(a[i]) { case 'a': suma+=10*convert(scale,j); break; case 'b': suma+=11*convert(scale,j); break; case 'c': suma+=12*convert(scale,j); break; case 'd': suma+=13*convert(scale,j); break; case 'e': suma+=14*convert(scale,j); break; case 'f': suma+=15*convert(scale,j); break; } } } return suma;}int main(){ int t; scanf("%d%*c",&t); char a[34],b[34]; int scale; while(t--) { scanf("%s %s %d",a,b,&scale); int suma=0,sumb=0; suma=account(a,suma,scale); sumb=account(b,sumb,scale); printf("(%d,%d)\n",suma/sumb,suma%sumb); } return 0;}
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