Codeforces Round #473 (Div. 2)
Codeforces Round #473 (Div. 2)
attack
发布于 2018-08-01 11:18:33
发布于 2018-08-01 11:18:33
这场怎么都是异或啊qwq。。。
A. Mahmoud and Ehab and the even-odd game
直接判断奇偶性
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
using namespace std;
const int MAXN = 2 * 1e6 + 10, INF = 1e9 + 10, B = 63;
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;
}
main() {
#ifdef WIN32
//freopen("a.in", "r", stdin);
#endif
int N = read();
puts(N & 1 ? "Ehab" : "Mahmoud");
}B. Mahmoud and Ehab and the message
维护出每个graph的最小值，输出即可，我咋还开了个map
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#define int long long
using namespace std;
const int MAXN = 2 * 1e6 + 10, INF = 1e9 + 10, B = 63;
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, G, M;
string s[MAXN];
int val[MAXN], minval[MAXN];
map<string, int> ID;
main() {
#ifdef WIN32
//freopen("a.in", "r", stdin);
#endif
memset(minval, 0x3f, sizeof(minval));
N = read(); G = read(); M = read();
for(int i = 1; i <= N; i++)
cin >> s[i];
//cout << s << endl;
for(int i = 1; i <= N; i++)
val[i] = read();
for(int i = 1; i <= G; i++) {
int num = read();
for(int j = 1; j <= num; j++) {
int x = read();
ID[s[x]] = i;
}
}
for(int i = 1; i <= N; i++)
minval[ID[s[i]]] = min(minval[ID[s[i]]], val[i]);
//for(int i = 1; i <= N; i++) printf("%d ", ID[s[i]]); puts("***");
//for(int i = 1; i <= N; i++) printf("%d ", minval[ID[s[i]]]); puts("***");
int ans = 0;
string x;
for(int i = 1; i <= M; i++) {
cin >> x;
ans += minval[ID[x]];
}
printf("%I64d", ans);
}C. Mahmoud and Ehab and the wrong algorithm
我的构造思路比较奇葩，大概长这样。。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
const int MAXN = 2 * 1e6 + 10, INF = 1e9 + 10, B = 63;
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;
}
main() {
#ifdef WIN32
//freopen("a.in", "r", stdin);
#endif
int N = read();
if(N == 2 || N == 3 || N == 4 || N == 5) printf("-1\n");
else {
printf("1 2\n");
printf("1 3\n");
printf("1 4\n");
for(int i = 5; i <= N; i++)
printf("3 %d\n", i);
}
for(int i = 1; i <= N - 1; i++)
printf("%d %d\n", i, i + 1);
}
E. Mahmoud and Ehab and the xor-MST
打表后强上oeis
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#define int long long
using namespace std;
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<int, int> ans;
int B(int x) {
if(ans[x] ) return ans[x];
if(x & 1) return ans[x] = 2 * B(x / 2) + (x / 2) + 1;
else return ans[x] = 2 * B(x / 2) + x / 2;
}
main() {
#ifdef WIN32
//freopen("a.in", "r", stdin);
#endif
ans[1] = 1;
int N = read();
printf("%I64d", B(N - 1));
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原始发表：2018-06-16 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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