HDU 1753 大明A+B

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1753

小数的高精度运算

思路 把小数点前后分开存，然后统一进位，输出时再输出小数点。

View Code

  1 #include <stdio.h>

  2 #include <string.h>

  3 #define N 405

  4 int a[N];

  5 int b[N];

  6 int c[N];

  7 int a1[N];

  8 int b1[N];

  9 int c1[N];

 10 int d[N];

 11 char stra[1000];

 12 char strb[1000];

 13 int max2(int q,int p)

 14 {

 15     return q>p?q:p;

 16 }

 17 int main()

 18 {

 19     int i,j;

 20     while (scanf("%s %s",stra,strb)!=EOF)

 21     {

 22         int alen = strlen (stra);

 23         int flaga=0,flagb=0;

 24         for (i=0;i<alen;i++)

 25         {

 26             if (stra[i] == '.')

 27             {

 28                 flaga = i;

 29                 break;

 30             }

 31         }

 32         if (i == alen)

 33         {

 34             flaga = alen;

 35         }

 36         int aplen ;

 37         int blen = strlen(strb);

 38         for (i=0;i<blen;i++)

 39         {

 40             if (strb[i] == '.')

 41             {

 42                 flagb = i;

 43                 break;

 44             }

 45         }

 46         if (i == blen)

 47         {

 48             flagb = blen;

 49         }

 50         int bplen ;

 51         memset (a,0,sizeof(a));

 52         memset (b,0,sizeof(b));

 53         memset (c,0,sizeof(c));

 54         memset (a1,0,sizeof(a1));

 55         memset (b1,0,sizeof(b1));

 56         memset (c1,0,sizeof(c1));

 57         j=0;

 58         for (i=flaga-1;i>=0;i--)

 59         {

 60             a[j] = stra[i]-'0';

 61             j++;

 62         }

 63         int aclen = j--;

 64         j=0;

 65         for (i=flagb-1;i>=0;i--)

 66         {

 67             b[j] = strb[i]-'0';

 68             j++;

 69         }

 70         int bclen = j--;

 71         j=1;

 72         for (i=flaga+1;i<alen;i++)

 73         {

 74             a1[j] = stra[i]-'0';

 75             j++;

 76         }

 77         aplen = j--;

 78         j=1;

 79         for (i=flagb+1;i<blen;i++)

 80         {

 81             b1[j] = strb[i]-'0';

 82             j++;

 83         }

 84         bplen = j--;

 85         int maxplen = max2(aplen,bplen);

 86         for (i=maxplen;i>=1;i--)

 87         {

 88             c1[i-1] = (a1[i]+b1[i]+c1[i])/10;

 89             c1[i] = (a1[i]+b1[i]+c1[i])%10;

 90 

 91         }

 92         int maxclen = max2(aclen,bclen);

 93         for (i=0;i<maxclen;i++)

 94         {

 95             c[i+1] = (a[i]+b[i]+c[i])/10;

 96             c[i] = (a[i]+b[i]+c[i])%10;

 97         }

 98         if (c1[0])

 99         {

100             memset(d,0,sizeof(d));

101             d[0] = 1;

102             for (i=0;i<maxclen;i++)

103             {

104                 c[i+1] += (c[i]+d[i])/10;

105                 c[i] = (c[i]+d[i])%10;

106             }

107         }

108         int flag = 0;

109         for (i=maxclen+1;i>=0;i--)

110         {

111             if(c[i]!=0)

112                 flag = 1;

113             if(flag)

114                 printf("%d",c[i]);

115         }

116         if(flag == 0)

117             printf("0");

118         flag = 0;

119         for (i=1;i<=maxplen;i++)

120         {

121             if(c1[i]!=0)

122                 flag = i;

123         }

124         if(flag)

125         {

126             printf(".");

127             for (i=1;i<=flag;i++)

128                 printf("%d",c1[i]);

129         }

130         printf("\n");

131     }

132     return 0;

133 }