C语言韩信点兵（容易超时） 求韩信点兵算法（c语言），限时500ms,求解？
c\u8bed\u8a00\u7f16\u7a0b \u97e9\u4fe1\u70b9\u5175 \u603b\u662f\u8d85\u65f6 \u7f51\u4e0a\u7684\u529e\u6cd5\u7528\u8fc7\u90fd\u4e0d\u884c\u6c42\u89e3 \u56f0\u6270\u6211\u597d\u4e45\u4e86 \u8fd8\u6709\u5982\u4f55\u5982\u4f55\u7528\u679a\u4f60\u7684\u56fe\u7247\u770b\u4e0d\u6e05\u695a\uff0c\u622a\u5c4f\u6216\u8005\u53d1\u94fe\u63a5\u3002
\u7b54\u6848\uff1a22439962
递归的思想，1个队列的话，3余1，明显就是3x+1，最小x=1；和为4两个队列，5余2,，首先他要满足4+3x的形式，才能满足1，找最小的x使3x+4%5=2；5次以内必找到，和为7；
三个队列7+(3*5)x%7=4；7次必找到；可以得出最多只要计算A+B+C+D+E+F+G+H次的乘法和除法，不可能超1000MS。
Sample Input
3 5 7 13 17 19 23 29
1 2 4 6 2 10 1 11
Sample Output
1306792
这个结果不对啊，1306792取余29是23，不是11！
用下面的程序，输出结果是224399662！
#include <stdio.h>
void main()
{
unsigned int i;
unsigned int col[8];
unsigned int rem[8];
unsigned long col_count;
unsigned int item;
unsigned long total;
while(1)
{
total = 0xffffffff;
printf("total(0xffffffff) = %lu\n",total);
printf("1: Start\n");
printf("0: Exit\n");
printf("Please select the operation:\n");
scanf("%d",&item);
if(item != 1)
break;
printf("Please input the condition:\n");
scanf("%d %d %d %d %d %d %d %d",&col[0],&col[1],&col[2],&col[3],&col[4],&col[5],&col[6],&col[7]);
scanf("%d %d %d %d %d %d %d %d",&rem[0],&rem[1],&rem[2],&rem[3],&rem[4],&rem[5],&rem[6],&rem[7]);
printf("Calculating,please waiting...\n");
printf("%d %d %d %d %d %d %d %d\n",col[0],col[1],col[2],col[3],col[4],col[5],col[6],col[7]);
printf("%d %d %d %d %d %d %d %d\n",rem[0],rem[1],rem[2],rem[3],rem[4],rem[5],rem[6],rem[7]);
item = 2;
col_count = 0;
while(1)
{
col_count ++;
total = col_count * col[7] + rem[7];
for(i=0; i<7; i++)
{
if((total % col[i]) != rem[i])
break;
}
if(i>=7)
break;
if(total >= (0xffffffff - col[7] - rem[7]))
{
printf("No result! Continue?(1:yes 0:no)");
scanf("%d",&item);
break;
}
}
if(item == 1)
continue;
else if(item == 0)
break;
else
{
printf("i = %d\n",i);
printf("The total army is: %lu\n",total);
printf("Continue?(1:yes 0:no)");
scanf("%d",&item);
if(item == 0)
break;
}
}
}
用c语言编程问题补充：最后答案是1051！ #include
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