UVa 10551 - Basic Remains

来源：互联网发布：js array map index 编辑：程序博客网时间：2024/06/03 13:48

题目：给你两个b进制数p，m，求p mod m的余数的b进制表示。

分析：数论，大整数。可以转成10进制在转回去，这里直接处理b进制。

处理过程和10进制相同，移位减法即可（借位是+b）。

说明：写的有点复杂╮(╯▽╰)╭。

#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>typedef struct _bn{int length;int data[1111];}bignumber;bignumber bigNumber(char buf[]){bignumber temp;temp.length = strlen(buf);for (int i = 0; buf[i]; ++ i)temp.data[i] = buf[temp.length-1-i]-'0';return temp;}void bigNumberOutput(bignumber A){int nowA = A.length-1;if (nowA == -1) printf("0");while (nowA >= 0)printf("%d",A.data[nowA --]);printf("\n");}bool biger(bignumber A, bignumber B){int nowA = A.length-1, nowB = B.length-1;while (nowB >= 0) {if (A.data[nowA] > B.data[nowB])return true;if (A.data[nowA] < B.data[nowB])return false;nowA --; nowB --;}return true;}bignumber bigNumberSub(bignumber A, bignumber B, int base){int nowA = A.length-1, nowB = B.length-1;while (nowB >= 0)A.data[nowA --] -= B.data[nowB --];while (nowA < A.length) {if (A.data[nowA] < 0) {A.data[nowA] += base;A.data[nowA+1] -= 1;}++ nowA;}while (A.length > 1 && !A.data[A.length-1])-- A.length;return A;}bignumber bigNumberMod(bignumber A, bignumber B, int base){while (A.length > B.length) {if (biger(A, B))A = bigNumberSub(A, B, base);else{A.data[A.length-2] += A.data[A.length-1]*base;A.data[A.length --] = 0;}}while (A.length == B.length && biger(A, B))A = bigNumberSub(A, B, base);return A;}int main(){int  base;char buf[1111];while (~scanf("%d",&base) && base) {scanf("%s",buf);bignumber P = bigNumber(buf);scanf("%s",buf);bignumber M = bigNumber(buf);bigNumberOutput(bigNumberMod(P, M, base));}    return 0;}

0 0