POJ2602 Superlong sums【水题】
Superlong sums Description The creators of a new programming language D++ have found out that whatever limit for SuperLongInt type they make,sometimes programmers need to operate even larger numbers. A limit of 1000 digits is so small... You have to find the sum of two numbers with maximal size of 1.000.000 digits. Input The first line of an input file contains a single number N (1<=N<=1000000) - the length of the integers (in order to make their lengths equal,some leading zeroes can be added). It is followed by these integers written in columns. That is,the next N lines contain two digits each,divided by a space. Each of the two given integers is not less than 1,and the length of their sum does not exceed N. Output Output file should contain exactly N digits in a single line representing the sum of these two integers. Sample Input 4 Sample Output 4750 Huge input,scanf is recommended. Source Ural State University collegiate programming contest 2000 问题链接:POJ2602 Superlong sums AC的C语言程序如下: /* POJ2602 Superlong sums */ #include <stdio.h> #define N 1000000 char a[N + 1]; int main(void) { int n,b,carry,i; scanf("%d",&n); for(i = 0; i < n; i++) { scanf("%d%d",&a[i],&b); a[i] += b; } for(carry = 0,i--; i >= 0; i--) { a[i] += carry; carry = a[i] / 10; a[i] %= 10; a[i] += '0'; } puts(a); return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |