求解所有的水仙花数（回归数）----大数字位数量级别的运算算法 - c++编程基础

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#include
#include
#include

#define base 100000000
#define ds 5

int powD[10][ds];
int sum[ds];
int num[10];
int N;
struct timeva l start, end;

void init(void)
{
int i, j, k;

for (i = 0; i <= 9; i ++)
for (j = 0; j < ds; j ++)
powD[i][j] = 0;

for (i = 0; i <= 9; i ++)
num[i] = 0;

for (i = 1; i <= 9; i ++)
powD[i][0] = i;
}

void nextPower(void)
{
int i, j, t;
for (i = 2; i <= 9; i ++)
{
t = 0;
for (j = 0; j < ds; j ++)
{
t = powD[i][j] * i + t;
if (t >= base)
{
powD[i][j] = t % base;
t /= base;
}
else
{
powD[i][j] = t;
t = 0;
}
}
}
}

int circle(int n, int b)
{
int i, j, k;
double tm;

if (n == 0)
{
if (sort())
{
k = ds - 1;
while ((sum[k] == 0) && (k >= 0))
k --;
gettimeofday(&end, NULL);
tm = (end.tv_sec - start.tv_sec) * 1000000.0 + (end.tv_usec - start.tv_usec);
tm /= 1000000.0;
printf("%4.4f ", tm);
printf("%2d: ", N);
printf("%d", sum[k]);
if (k > 0)
for (i = k - 1; i >= 0; i --)
printf("%08d", sum[i]);
printf("\n");
}
}
else
for (i = 0; i <= b; i ++)
{
num[i] ++;
circle(n-1, i);
num[i] --;
}
}

int sort(void)
{
int i, j, t;
int temp[ds], d[10];

for (i = 0; i < ds; i ++)
temp[i] = sum[i] = 0;

for (i = 0; i <= 9; i ++)
d[i] = 0;

for (i = 1; i <= 9; i ++)
if (num[i])
{
t = 0;
for (j = 0; j < ds; j ++)
{
t = powD[i][j] * num[i] + t;
if (t >= base)
{
temp[j] = t % base;
t /= base;
}
else
{
temp[j] = t;
t = 0;
}
}

t = 0;
for (j = 0; j < ds; j++)
{

t = temp[j] + sum[j] + t;
if (t >= base)
{
sum[j] = t - 100000000;
t = 1;
}
else
{
sum[j] = t;
t = 0;
}
}
}

for (i = ds - 1; i >= 0; i --)
{
t = sum[i];
if (t > 0)
while (t > 0)
{
d[t % 10] ++;
t /= 10;
}
}

for (i = 1; i <= 9; i ++)
if (num[i] != d[i]) return 0;
return 1;
}

int main(void)
{
int i, j;

gettimeofday(&start, NULL);
init();
nextPower(); //是平方
for (i = 2; i <= 40; i ++)
{
N = i;
printf("Search %d...\n", i);
circle(N, 9);
nextPower();
}
return 0;
}

#include
#include
#include

#define base 100000000
#define ds 5

int powD[10][ds];
int sum[ds];
int num[10];
int N;
struct timeva l start, end;

void init(void)
{
int i, j, k;

for (i = 0; i <= 9; i ++)
for (j = 0; j < ds; j ++)
powD[i][j] = 0;

for (i = 0; i <= 9; i ++)
num[i] = 0;

for (i = 1; i <= 9; i ++)
powD[i][0] = i;
}

void nextPower(void)
{
int i, j, t;
for (i = 2; i <= 9; i ++)
{
t = 0;
for (j = 0; j < ds; j ++)
{
t = powD[i][j] * i + t;
if (t >= base)
{
powD[i][j] = t % base;
t /= base;
}
else
{
powD[i][j] = t;
t = 0;
}
}
}
}

int circle(int n, int b)
{
int i, j, k;
double tm;

if (n == 0)
{
if (sort())
{
k = ds - 1;
while ((sum[k] == 0) && (k >= 0))
k --;
gettimeofday(&end, NULL);
tm = (end.tv_sec - start.tv_sec) * 1000000.0 + (end.tv_usec - start.tv_usec);
tm /= 1000000.0;
printf("%4.4f ", tm);
printf("%2d: ", N);
printf("%d", sum[k]);
if (k > 0)
for (i = k - 1; i >= 0; i --)
pri

求解所有的水仙花数（回归数）----大数字位数量级别的运算算法 (一)