河南省第七届省赛 问题 F: Turing equation 水题

**题目描述
The fight goes on, whether to store numbers starting with their most significant digit or their least significant digit. Sometimes this is also called the “Endian War”. The battleground dates far back into the early days of computer science. Joe Stoy, in his (by the way excellent) book “Denotational Semantics”, tells following story:

“The decision which way round the digits run is, of course, mathematically trivial. Indeed, one early British computer had numbers running from right to left (because the spot on an oscilloscope tube runs from left to right, but in serial logic the least significant digits are dealt with first). Turing used to mystify audiences at public lectures when, quite by accident, he would slip into this mode even for decimal arithmetic, and write things like 73+42=16. The next version of the machine was made more conventional simply by crossing the x-deflection wires: this, however, worried the engineers, whose waveforms were all backwards. That problem was in turn solved by providing a little window so that the engineers (who tended to be behind the computer anyway) could view the oscilloscope screen from the back.

You will play the role of the audience and judge on the truth value of Turing’s equations.

输入
The input contains several test cases. Each specifies on a single line a Turing equation. A Turing equation has the form “a+b=c”, where a, b, c are numbers made up of the digits 0,…,9. Each number will consist of at most 7 digits. This includes possible leading or trailing zeros. The equation “0+0=0” will finish the input and has to be processed, too. The equations will not contain any spaces.

输出
For each test case generate a line containing the word “TRUE” or the word “FALSE”, if the equation is true or false, respectively, in Turing’s interpretation, i.e. the numbers being read backwards.**
这道题就是，输入等式后 把等式中的各个整数颠倒后再进行相加，判断是否成立，且以“0+0=0”作为结束。
例如输入61+51=13输出TRUE，因为将各个数字颠倒后为16+15=31，该式子成立
用find标记这是第几个数，从后往前判断，若find 为1，则为第一个数，若find 为2，则为第二个数，若find为3,则为第三个数，在最后判断一下，若第二个数加第三个数等于第一个数 则输出true。从后往前遍历时，记得跳过前导0

#include<stdio.h> #include<string.h>char str[100];int main() {int l,i,sum,find,d1,d2,d3;while(scanf("%s",str)!=EOF){l = strlen(str);if(strcmp(str,"0+0=0")==0)break;while(str[l-1] == '0')l--;if(str[l-1] == '=')l++;find = 0;sum = 0;for( i = l-1; i >= 0;i--){sum = 0;while(str[i]>= '0'&&str[i]<='9'){sum = sum*10 + str[i]-'0';i--;}if( sum >= 0)find ++;if(find == 1)d1 = sum;if(find == 2)d2 = sum;if(find == 3)d3 = sum;}if( d2 + d3 == d1)printf("TRUE\n");elseprintf("FALSE\n");}return 0; }

转载于:https://www.cnblogs.com/hellocheng/p/7350169.html

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