【应用类型源码】【记忆翻牌游戏源码】【eclipse 无法查看源码】c 电话 源码_电话簿源码
1.手机通讯录管理中拨号的话源话簿源代码c语言编程
2.用C语言求1234567890123456789+9876543210123456789，最好源代码有解释，码电万分感谢
手机通讯录管理中拨号的源码应用类型源码源代码c语言编程
IplImage* cameraImage = NULL;
while ((cameraImage = cvQueryFrame(capture)) != NULL)
{
cvShowImage("Camera",cameraImage);
cvWaitKey(1);
}
用C语言求+，最好源代码有解释，话源话簿万分感谢
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char a[][] = { 0 };
char result[] = { '0' };
int n,码电记忆翻牌游戏源码 i, j;
scanf("%d", &n);
for (i = 0; i < n; ++i)
scanf("%s", a[i]);
for (i = 0; i < n; ++i)
{
int len1 = strlen(a[i]);
int len2 = strlen(result);
if (len1 < len2)
{
int temp = 0;
for (j = len2 - 1; j >= len2 - len1; --j)
{
if (temp + result[j] + a[i][len1 + j - len2] - '0' - '0' >= )
{
result[j] = temp + result[j] + a[i][len1 + j - len2] - '0' - ;
temp = 1;
}
else
{
result[j] = temp + result[j] + a[i][len1 + j - len2] - '0';
temp = 0;
}
}
for (j = len2 - len1 - 1; j >= 1; --j)
{
if (result[j] + temp <= '9')
{
result[j] += temp;
temp = 0;
}
else
{
result[j] = '0';
temp = 1;
}
}
if (result[0] + temp <= '9')result[0] += temp;
else
{
result[0] = '0';
for (j = len2 - 1; j >= 0; --j)
result[j + 1] = result[j];
result[0] = '1';
}
}
else
{
int temp = 0;
char TEMP[] = { 0 };
for (j = len1 - 1; j >= len1 - len2; --j)
{
if (temp + a[i][j] + result[len2 + j - len1] - '0' - '0' >= )
{
TEMP[j] = temp + result[len2 + j - len1] + a[i][j] - '0' - ;
temp = 1;
}
else
{
TEMP[j] = temp + result[len2 + j - len1] + a[i][j] - '0';
temp = 0;
}
}
for (j = len1 - len2 - 1; j >= 1; --j)
{
if (a[i][j] + temp <= '9')
{
TEMP[j] = temp + a[i][j];
temp = 0;
}
else
{
TEMP[j] = '0';
temp = 1;
}
}
if (len1 > len2)
{
if (a[i][0] + temp <= '9')TEMP[0] = temp + a[i][0];
else
{
TEMP[0] = '0';
for (j = len1 - 1; j >= 0; --j)
TEMP[j + 1] = TEMP[j];
TEMP[0] = '1';
}
}
else
{
if (temp == 1)
{
for (j = len1 - 1; j >= 0; --j)
TEMP[j + 1] = TEMP[j];
TEMP[0] = '1';
}
}
strcpy(result, TEMP);
}
}
int zero = 0;
while (result[zero] == '0')
{
++zero;
}
int length = strlen(result);
for (j = zero; j < length; ++j)
result[j - zero] = result[j];
length -= zero;
for (j = length; j < ; ++j)
result[j] = 0;
if (result[0] == 0)result[0] = '0';
printf("%s", result);
system("pause");
return 0;
}
以前做的一个，可以求多个大数的源码eclipse 无法查看源码和。你这个问题就属于求大数的话源话簿和。关键思路是码电字符串化，难点在于进位，源码用一个temp储存进位与否即可。话源话簿
码电码电
