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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#450851 | #1289. A + B Problem | Froranzen | AC ✓ | 25ms | 16868kb | C++20 | 2.4kb | 2024-06-22 18:36:00 | 2024-06-22 18:36:01 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i, f, t) for(int i(f); i <= t; ++i)
#define re(i, t) for(int i(1); i <= t; ++i)
#define per(i, t, f) for(int i(t); i >= f; --i)
#define pe(i, t) for(int i(t); i >= 1; --i)
#define nx(i, u) for(int i(head[u]); i; i = e[i].nxt)
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
typedef pair <int, int> pii;
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define Ls (ix(l, mid))
#define Rs (ix(mid+1, r))
#define ix(l, r) ((l + r) | (l != r))
#define mp(i, j) (make_pair(i, j))
//#define int long long
#define inf 0x3f3f3f3f
#define i128 __int128
#define INF 0x3f3f3f3f3f3f3f3f
namespace IO {
char buf[1 << 21], *p1 = buf, *p2 = buf, buf1[1 << 21];
inline char gc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
template<class I>
inline void read(I &x) {x = 0;I f = 1;char c = gc();while (c < '0' || c > '9') {if (c == '-') {f = -1;} c = gc();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc();}x *= f;}
template<class I>
inline void write(I x) {if (x == 0) {putchar('0');return;}I tmp = x > 0 ? x : -x;if (x < 0) {putchar('-');}int cnt = 0;while (tmp > 0) {buf1[cnt++] = tmp % 10 + '0';tmp /= 10;}while (cnt > 0)putchar(buf1[--cnt]);}
#define outn(x) write(x), putchar('\n')
#define out(x) write(x), putchar(' ')
} using namespace IO;
#define look_memory cerr<<abs(&sT-&eD)/1024.0/1024<<'\n'
int sT;
const int N = 4e6 + 5;
int T, n, m;
char s[N];
int a[N], b[N];
int pos[N], ax[N], ay[N];
void solve () {
if(n < m) swap(n, m);
int len = n + m + 2;
int L = 0;
int sz = 0;
re(i, m) {
sz += (!a[i]);
if(a[i]) {
pos[++L] = m-i+1;
ay[m-i+1] = 1;
}
}
rep(i, m+1, n+m) {
if(a[i]) ax[n-(i-m)+1] = 1;
}
rep(i, m+1, n+m) {
if(!a[i]) {
++sz;
if(sz > m) break;
if(pos[L] > (n-(i-m)+1)) break;
ay[pos[L]] = 0;
--L;
ax[n-(i-m)+1] = 1;
}
}
re(i, len) {
b[i] = ax[i] + ay[i];
}
re(i, len) {
b[i+1] += b[i]/2;
b[i] &= 1;
}
len = 1;
re(i, n+m+2) {
if(b[i]) len = i;
}
pe(i, len) {
if(b[i]) putchar('1');
else putchar('0');
}
puts("");
re(i, n+m+2) {
ax[i] = ay[i] = b[i] = 0;
}
}
int main (){
scanf("%d", &T);
while(T--) {
scanf("%d %d %s", &n, &m, s + 1);
re(i, n+m) a[i] = s[i] - '0';
solve();
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 15932kb
input:
3 4 3 1000101 2 2 1111 1 1 00
output:
1101 110 0
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 25ms
memory: 16868kb
input:
11110 10 8 111011010011100100 3 5 01011000 7 6 1110101010000 9 1 0110100101 1 9 0100001110 8 10 000101101011111000 9 6 011111111000111 1 9 1011101101 10 7 00100011000100000 4 9 1000101101010 8 4 100100110000 8 9 00101111011000101 8 9 11000000101011110 7 6 1111010100110 2 9 01001110101 4 5 100010100 ...
output:
10011010100 11100 10101000 110100101 100001110 10000001100 1000010111 111101101 1110100000 111101010 11110000 1000011101 1001011110 10101110 101110101 11100 1111010 1000010 1011100010 10010101001 10010001 1001010 1000000010 1110 111 1111110001 10110111 1100010101 10000000 111000011 110 11111 1100101...
result:
ok 11110 lines