時間限制?1000 ms?內(nèi)存限制?65536 KB

題目描述

?? ?Hawei is learning C programming language recently, but he is so naive that he mistakes the?～?symbol in C for the???symbol in mathemetics. In math calculation, the answer to??4?is the bitwise NOT of?(100)2, ie.,??(4)10=(011)2=(3)10. Thus,??4=3. But in C programming, the compiler will explain calculation?ans=～4?as the bitwise NOT of?(00000000000000000000000000000100)2?because?4?is a 32-bit?integer,?ie.,?ans=～(00000000000000000000000000000100)2=(11111111111111111111111111111011)2?=(?5)10.

? ??Unaware of this slight difference, Hawei checked for the bug for a whole day. When he finally found it, he was so angry that he decided?to define two new functions?f(x)?and?G(x). Function?f(x)?returns the answer of??x, in other words, it is the bitwise NOT of?xsince the most significant digit of?x. For instance, we have:?
?? ?f((4)10)=f((100)2)=(011)2=(3)10
?? ?, and
?? ?f((22)10)=f((10110)2)=(01001)2=(9)10

?? ?On the other hand, function?G(x)?is defined in following way:

輸入格式

?? ?The input starts with a single line containing an integer?T?(1≤T≤10), indicating the number of test cases.

For each test case, the first line contains one integers?N(1≤N≤100000), indicating the length?of the string. The next lines contains a string?S, representing the integer?x?of the corresponding test case in its binary form. It is guaranteed that there is no leading zeroes in the input.

?

輸出格式

?? ?For each test case, output a single line containing a 01 string, indicating the answer of?G(x)?in binary form. No leading zeroes are allowed.

輸入樣例

1 3 100

輸出樣例

1 找規(guī)律，統(tǒng)計輸入串中1的數(shù)量然后輸出相應(yīng)數(shù)量的1即可

#include<cstdio> #define N 100050 using namespace std; int main(){int t,n,i;char str[N];for(scanf("%d",&t);t--;){scanf("%d",&n);scanf("%s",str);for(i=0;str[i];i++){if(str[i]=='1'){printf("1");}}printf("\n");}return 0; }

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