标签:print put math.h scan 题目 找规律 this red down
第1行:一个数T,表示后面用作输入测试的数的数量。(1 <= T <= 1000) 第2 - T + 1行:每行1个数N。(1 <= N <= 10^1000)
共T行,如果A获胜输出A,如果B获胜输出B。
3 2 3 4
A B A
2的幂看上去很厉害的样子,然而这题还是属于bash博弈的变形,所以做法还是打表找规律
然后发现只要是3的倍数就是B赢
不过这题有个问题,就是会有大数,然后我就直接套了个大数模板 其实想起来只要每一位去除3看可不可以就行了 不过也懒得改了
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;
const int MX = 2500;
const int MAXN = 9999;
const int DLEN = 4;
/*已重载>+- % 和print*/
class Big {
public:
int a[MX], len;
Big(const int b = 0) {
int c, d = b;
len = 0;
memset(a, 0, sizeof(a));
while(d > MAXN) {
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
Big(const char *s) {
int t, k, index, L, i;
memset(a, 0, sizeof(a));
L = strlen(s);
len = L / DLEN;
if(L % DLEN) len++;
index = 0;
for(i = L - 1; i >= 0; i -= DLEN) {
t = 0;
k = i - DLEN + 1;
if(k < 0) k = 0;
for(int j = k; j <= i; j++) {
t = t * 10 + s[j] - ‘0‘;
}
a[index++] = t;
}
}
Big operator/(const int &b)const {
Big ret;
int i, down = 0;
for(int i = len - 1; i >= 0; i--) {
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
bool operator>(const Big &T)const {
int ln;
if(len > T.len) return true;
else if(len == T.len) {
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0) ln--;
if(ln >= 0 && a[ln] > T.a[ln]) return true;
else return false;
} else return false;
}
Big operator+(const Big &T)const {
Big t(*this);
int i, big;
big = T.len > len ? T.len : len;
for(i = 0; i < big; i++) {
t.a[i] += T.a[i];
if(t.a[i] > MAXN) {
t.a[i + 1]++;
t.a[i] -= MAXN + 1;
}
}
if(t.a[big] != 0) t.len = big + 1;
else t.len = big;
return t;
}
Big operator-(const Big &T)const {
int i, j, big;
bool flag;
Big t1, t2;
if(*this > T) {
t1 = *this;
t2 = T;
flag = 0;
} else {
t1 = T;
t2 = *this;
flag = 1;
}
big = t1.len;
for(i = 0; i < big; i++) {
if(t1.a[i] < t2.a[i]) {
j = i + 1;
while(t1.a[j] == 0) j++;
t1.a[j--]--;
while(j > i) t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
} else t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1) {
t1.len--;
big--;
}
if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];
return t1;
}
int operator%(const int &b)const {
int i, d = 0;
for(int i = len - 1; i >= 0; i--) {
d = ((d * (MAXN + 1)) % b + a[i]) % b;
}
return d;
}
Big operator*(const Big &T) const {
Big ret;
int i, j, up, temp, temp1;
for(i = 0; i < len; i++) {
up = 0;
for(j = 0; j < T.len; j++) {
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN) {
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
} else {
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0) {
ret.a[i + j] = up;
}
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
void print() {
printf("%d", a[len - 1]);
for(int i = len - 2; i >= 0; i--) printf("%04d", a[i]);
}
};
int main() {
//FIN
int T;
scanf("%d", &T);
char s[1005];
while(T--) {
scanf("%s", s);
Big B(s);
//B.print();
if(B % 3 != 0) printf("A\n");
else printf("B\n");
}
return 0;
}
标签:print put math.h scan 题目 找规律 this red down
原文地址:http://www.cnblogs.com/Hyouka/p/7368024.html