HDU-1753 大明A+B【大数相加】

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

解题思路：

大数相加问题，但是有三种情况：2个整数，1个整数1个小数，2个小数。

这就需要我们单独处理每种情况，而且这道题需要保证没有多余的前导0和后导0，刚开始想到简单了。结果到最后搞了150行才过。悲剧。

这道题的巧妙做法是：把每种情况都转换成2个小数来搞，这样就只有1种情况了。

代码如下：

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int N = 810;char a[N], b[N];int c[N], d[N];int main(){int flag1, flag2, len1, len2, pos;int up, top, down;while(scanf("%s %s", a, b) != EOF){flag1 = flag2 = -1;memset(c, 0, sizeof(c));memset(d, 0, sizeof(d));len1 = strlen(a); len2 = strlen(b);for(int i = 0; i < len1; ++i) //整数部分、小数部分if(a[i] == '.') flag1 = i;for(int i = 0; i < len2; ++i) //整数部分、小数部分if(b[i] == '.') flag2 = i;if(flag1 == - 1 && flag2 == -1) //2整{for(int i = len1 - 1, k = 0; i >= 0; --i)c[k++] = a[i] - '0';for(int i = len2 - 1, k = 0; i >= 0; --i)d[k++] = b[i] - '0';for(int i = 0, up = 0; i < N; ++i){c[i] = c[i] + d[i] + up;up = c[i] / 10;c[i] %= 10;}for(top = N -1; !c[top] && top >= 0; --top);if(top < 0){printf("0\n");continue;}for(int i = top; i >= 0; --i)printf("%d", c[i]);printf("\n");}else if(flag1 != -1 && flag2 != -1) //2小{if(len1 - 1 - flag1 > len2 - 1 - flag2) //a小数部分长{for(int i = 0; i < len1 - len2 + flag2 - flag1; ++i)b[len2 + i] = '0';b[len1 + flag2 - flag1 + 1] = '\0';len2 += len1 - len2 + flag2 - flag1;}else if((len1 - 1 - flag1 < len2 - 1 - flag2)){for(int i = 0; i < len2 - len1 + flag1 - flag2; ++i)a[len1 + i] = '0';a[len2 + flag1 - flag2 + 1] = '\0';len1 += len2 - len1 + flag1 - flag2;}pos = 0;for(int i = len1 - 1; i >= 0; --i)if(a[i] != '.')pos++;elsebreak;for(int i = len1 - 1, k = 0; i >= 0; --i)if(a[i] != '.')c[k++] = a[i] - '0';for(int i = len2 - 1, k = 0; i >= 0; --i)if(b[i] != '.')d[k++] = b[i] - '0';for(int i = 0, up = 0; i < N; ++i){c[i] = c[i] + d[i] + up;up = c[i] / 10;c[i] = c[i] % 10;}for(int i = 0; i < N; ++i)a[i] = c[i] + '0';for(top = N - 1; a[top] == '0'; --top); //去前导0if(top < 0){printf("0\n");continue;}for(down = 0; a[down] == '0' && down < pos; ++down); //去后导0for(int i = top; i >= down; --i)if(i == pos && down != pos)printf("%c.", a[i]);elseprintf("%c", a[i]);printf("\n");}else //1整、1小{if(flag1 == -1) //b小{for(int i = len1 - 1, k = 0; i >= 0; --i)c[k++] = a[i] - '0';for(int i = flag2 - 1, k = 0; i >= 0; --i)d[k++] = b[i] - '0';for(int i = 0, up = 0; i < N; ++i){c[i] = c[i] + d[i] + up;up = c[i] / 10;c[i] %= 10;}for(top = N - 1; !c[top]; --top); //去前导0if(top < 0){printf("0\n");continue;}for(down = len2 - 1; b[down] == '0'; --down); //去后导0for(int i = top; i >= 0; --i)printf("%d", c[i]);if(b[down] != '.')for(int i = flag2; i <= down; ++i)printf("%c", b[i]);printf("\n");}else //a小{for(int i = flag1 - 1, k = 0; i >= 0; --i)c[k++] = a[i] - '0';for(int i = len2 - 1, k = 0; i >= 0; --i)d[k++] = b[i] - '0';for(int i = 0, up = 0; i < N; ++i){c[i] = c[i] + d[i] + up;up = c[i] / 10;c[i] %= 10;}for(top = N - 1; !c[top]; --top); //去前导0if(top < 0){printf("0\n");continue;}for(down = len1 - 1; a[down] == '0'; --down); //去后导0for(int i = top; i >= 0; --i)printf("%d", c[i]);if(a[down] != '.')for(int i = flag1; i <= down; ++i)printf("%c", a[i]);printf("\n");}}}return 0;}

JAVA则超级轻松水过啊！！！！！！！！150行和20行，郁闷啊。。。。

JAVA，你太强悍了。ORZ

代码如下：

import java.math.BigDecimal;import java.util.*;public class Main{public static void main(String[] args){BigDecimal a, b, c;Scanner cin = new Scanner(System.in);while(cin.hasNext()){a = cin.nextBigDecimal();b = cin.nextBigDecimal();c = a.add(b);if(c.compareTo(BigDecimal.ZERO) == 0)System.out.println("0");elseSystem.out.println(c.stripTrailingZeros().toPlainString());}}}