AtCoderBeginnerContest109题解
AtCoderBeginnerContest109题解
attack
发布于 2018-09-17 15:39:38
发布于 2018-09-17 15:39:38
第一次AK，真爽qwq
A
很zz啊，，直接判断三种情况就行
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int check(int a, int b, int c) {
return a * b * c & 1;
}
main() {
int A = read(), B = read();
if(check(1, A, B) || check(2, A, B) || check(3, A, B)) puts("Yes");
else puts("No");
return 0;
}
/*
2 2 1
1 1
2 1 1
*/B
直接模拟即可
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
map<string, bool> mp;
int N;
string s[MAXN];
main() {
int N = read();
for(int i = 1; i <= N; i++) {
cin >> s[i];
if(mp.find(s[i]) != mp.end()) {puts("No"); return 0;}
if(i > 1) {
int l = s[i - 1].length();
if(s[i][0] != s[i - 1][l - 1]) {puts("No"); return 0;
}
}
mp[s[i]] = 1;
}
puts("Yes");
return 0;
}
/*
2 2 1
1 1
2 1 1
*/C
一开始想二分来着，然后发现我zz了。
直接输出 所有距离与起始距离的最大公约数即可
证明显然。。。
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, a[MAXN];
main() {
int N = read(), X = read();
for(int i = 1; i <= N; i++) a[i] = read();
int g = abs(X - a[1]);
for(int i = 2; i <= N; i++) {
int dis = abs(X - a[i]);
g = __gcd(g, dis);
}
printf("%d", g);
return 0;
}
/*
2 2 1
1 1
2 1 1
*/D
一开始读错题了，我以为移动几个都可以，事实上只能移动一个qwq，而且我以为0不统计入答案
我还特地问了一下，真没想到他居然知道我问的什么
考虑一个很显然的正确做法。。
先把所有奇数点往下移动，直到移动到最后一行
再把最后一行的奇数点从左往右移动。。
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 501, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M;
int a[MAXN][MAXN];
int xx[6] = {0, -1, +1, 0, 0};
int yy[6] = {0, 0, 0, -1, +1};
int ans[MAXN * MAXN][7], cnt = 0;
void add(int i, int j, int wx, int wy) {
ans[++cnt][0] = i;
ans[cnt][1] = j;
ans[cnt][2] = wx;
ans[cnt][3] = wy;
}
main() {
N = read(); M = read();
for(int i = 1; i <= N; i++)
for(int j = 1; j <= M; j++)
a[i][j] = read();
for(int i = 1; i < N; i++) {
for(int j = 1; j <= M; j++) {
if(a[i][j] & 1) {
int wx = i + 1, wy = j;
a[i][j]--; a[wx][wy]++;
add(i, j, wx, wy);
}
}
}
for(int i = 1; i < M; i++) {
if(a[N][i] & 1) {
a[N][i]--; a[N][i + 1]++;
add(N, i, N, i + 1);
}
}
printf("%d\n", cnt);
for(int i = 1; i <= cnt; i++)
printf("%d %d %d %d\n", ans[i][0], ans[i][1], ans[i][2], ans[i][3]);
return 0;
}
/*
2 2 1
1 1
2 1 1
*/本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2018-09-09 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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