**題目描述
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.

輸入
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.

輸出
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.**
這道題就是，輸入等式后 把等式中的各個(gè)整數(shù)顛倒后再進(jìn)行相加，判斷是否成立，且以“0+0=0”作為結(jié)束。
例如輸入61+51=13輸出TRUE，因?yàn)閷⒏鱾€(gè)數(shù)字顛倒后為16+15=31，該式子成立
用find標(biāo)記這是第幾個(gè)數(shù)，從后往前判斷，若find 為1，則為第一個(gè)數(shù)，若find 為2，則為第二個(gè)數(shù)，若find為3,則為第三個(gè)數(shù)，在最后判斷一下，若第二個(gè)數(shù)加第三個(gè)數(shù)等于第一個(gè)數(shù) 則輸出true。從后往前遍歷時(shí)，記得跳過前導(dǎo)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; }

轉(zhuǎn)載于:https://www.cnblogs.com/hellocheng/p/7350169.html

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