UVa 10551 - Basic Remains

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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;
}

UVa 10551 - Basic Remains

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原文地址：http://blog.csdn.net/mobius_strip/article/details/46535709