if(n

只可能有1或0两种情况。因为相邻两个数可以抵消为1或者-1。于是判断n的奇偶，然后判断n div 2之后的奇偶，总共4种情况。这四种情况分别对应1或者0中的一种。而由于n很大，不能直接进行这个判断，于是需要通过判断n mod 4 之后的值来实现上述判断。这里值得注意的是，判断一个数mod 4之后的值，只需要考虑后面两位就可以，因为整个数可以表示为前面的那部分*100+后两位组成的两位数，其中100 mod 4=0。这样，这道数论题就解决了！

吃亏了,我用STRING,真的亏了!!!!

very easy

找规律

读入要用ansistring

水水水
测试数据 #0: Accepted, time = 15 ms, mem = 560 KiB, score = 20
测试数据 #1: Accepted, time = 0 ms, mem = 560 KiB, score = 20
测试数据 #2: Accepted, time = 0 ms, mem = 560 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 560 KiB, score = 20
测试数据 #4: Accepted, time = 0 ms, mem = 560 KiB, score = 20
c++
#include<cstdio>
using namespace std;
int main()
{
long long n;
scanf("%d",&n);//错误的输入居然能过，如果改成“%lld”不能过，必须用%d才能过，，，，醉了
if(n%4==1||n%4==2)printf("1");
if(n%4==3||n%4==0)printf("0");
return 0;
}


foo.cpp:5:6: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
main()
^
测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 20
测试数据 #1: Accepted, time = 0 ms, mem = 560 KiB, score = 20
测试数据 #2: Accepted, time = 0 ms, mem = 564 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 560 KiB, score = 20
测试数据 #4: Accepted, time = 0 ms, mem = 560 KiB, score = 20
Accepted, time = 0 ms, mem = 564 KiB, score = 100
0
second

P1141最小非负值Accepted记录信息
评测状态 Accepted
题目 P1141 最小非负值
递交时间 2016-03-23 07:21:20
代码语言 C++
评测机 ShadowShore
消耗时间 0 ms
消耗内存 556 KiB
评测时间 2016-03-23 07:21:21

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 20

测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 20

测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 20

测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 20

测试数据 #4: Accepted, time = 0 ms, mem = 552 KiB, score = 20

Accepted, time = 0 ms, mem = 556 KiB, score = 100

我会告诉你我看到这道题的第一想法是DP么……

直到我看见了1e1000的数据范围……

var

        s:ansistring;

        j:longint;

begin

        readln(s);

        s:=copy(s,length(s)-1,length(s));

        val(s,j);

        j:=j mod 4;

        if j=0 then writeln(0) else

        if j=1 then writeln(1) else

        if j=2 then writeln(1) else

        if j=3 then writeln(0);

end.