2015-2016 ACM-ICPC, NEERC, Northern Subregional Contest (7/12)
作者：互联网
\[2015-2016\ ACM-ICPC,\ NEERC,\ Northern\ Subregional\ Contest\]
\(A.Alex\ Origami\ Squares\)
签到
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
int main(){
int a,b;
freopen("alex.in","r",stdin);
cin >> a >> b;
freopen("alex.out","w",stdout);
double res = max(max(min(a/3.0,b/1.0),min(a/1.0,b/3.0)),min(a/2.0,b/2.0));
cout << res << endl;
return 0;
}\(B.Black\ and\ White\)
构造，把多的框在少的里面即可
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
int n,m,len;
char tile[MAXN][5];
int main(){
#ifdef ONLINE_JUDGE
freopen("black.in","r",stdin);
#endif
____();
cin >> n >> m;
#ifdef ONLINE_JUDGE
freopen("black.out","w",stdout);
#endif
if(m==0){
cout << 1 << ' ' << 1 << endl << '@' << endl;
return 0;
}
if(n==0){
cout << 1 << ' ' << 1 << endl << '.' << endl;
return 0;
}
char x='@',y='.';
if(n<m){
swap(n,m);
swap(x,y);
}
len = 1;
if(n>m){
tile[len][1] = tile[len][2] = tile[len][3] = y;
len++, m--;
}
while(n>m){
tile[len][2] = x;
tile[len][1] = tile[len][3] = y;
len++, n--;
tile[len][1] = tile[len][2] = tile[len][3] = y;
len++;
}
while(n){
tile[len][1] = tile[len][2] = tile[len][3] = x;
len++, n--;
tile[len][1] = tile[len][2] = tile[len][3] = y;
len++;
}
if(tile[len][1]=='\0') len--;
cout << len << ' ' << 3 << endl;
for(int i = 1; i <= len; i++) cout << (tile[i]+1) << endl;
return 0;
}\(C.Concatenation\)
答案就是总数量-右串从\(1\)到\(lenr-1\)各个字符在左串\(2~lenl\)之间出现的次数
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 3e5+7;
char s[MAXN],t[MAXN];
int len1,len2;
int main(){
#ifdef ONLINE_JUDGE
freopen("concatenation.in","r",stdin);
#endif
scanf("%s %s",s+1,t+1);
#ifdef ONLINE_JUDGE
freopen("concatenation.out","w",stdout);
#endif
len1 = strlen(s+1);
len2 = strlen(t+1);
int_fast64_t res = len1*(int_fast64_t)len2;
map<char,int> mp;
for(int i = 2; i <= len1; i++) mp[s[i]]++;
for(int i = 1; i < len2; i++) res -= mp[t[i]];
cout << res << endl;
return 0;
}\(D.Distribution\ in\ Metagonia\)
考虑构造时先把质因子\(2\)，\(3\)提出来，然后得到一个奇数，把奇数拆分为\(3^k·p\)，继续对\(p\)拆分直到\(p\)的因子只含\(2\)和\(3\)为止
递归构造即可
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
int t;
LL n[1111];
void gao(LL m, LL mul, vector<LL> &vec){
if(m==1){
vec.emplace_back(mul);
return;
}
if(m%3!=0&&m%2!=0){
LL tp = 1;
while(tp*3<=m) tp *= 3;
gao(tp,mul,vec);
gao(m-tp,mul,vec);
return;
}
LL sum = 1;
while(m%2==0){
sum *= 2;
m >>= 1;
}
while(m%3==0){
sum *= 3;
m /= 3;
}
gao(m,sum*mul,vec);
}
void solve(LL m){
vector<LL> vec;
gao(m,1,vec);
cout << vec.size() << endl;
for(LL x : vec) cout << x << ' '; cout << endl;
}
int main(){
#ifdef ONLINE_JUDGE
freopen("distribution.in","r",stdin);
freopen("distribution.out","w",stdout);
#endif
____();
cin >> t;
for(int i = 1; i <= t; i++) cin >> n[i];
for(int i = 1; i <= t; i++) solve(n[i]);
return 0;
}\(E.Easy\ Arithmetic\)
打记号标记前导零、上一位是否是符号、当前正负值然后判断即可
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 3333;
char s[MAXN],tar[MAXN];
int main(){
#ifdef ONLINE_JUDGE
freopen("easy.in","r",stdin);
#endif
____();
cin >> (s+1);
int len = strlen(s+1);
#ifdef ONLINE_JUDGE
freopen("easy.out","w",stdout);
#endif
int c = 1, ptr = 1;
bool tag = true, lead = true, op = true;
while(ptr<=len){
if(!isdigit(s[ptr])){
if(s[ptr]=='-') tag = false;
else if(s[ptr]=='+') tag = true;
tar[c++] = s[ptr];
op = true;
}
else{
if(op){
op = false;
tar[c++] = s[ptr];
lead = (s[ptr]=='0');
}
else{
if(!tag){
tar[c++] = '+';
tag = true;
tar[c++] = s[ptr];
lead = (s[ptr]=='0');
}
else{
if(lead){
tar[c++] = '+';
lead = (s[ptr]=='0');
tar[c++] = s[ptr];
}
else tar[c++] = s[ptr];
}
}
}
ptr++;
}
cout << (tar+1) << endl;
return 0;
}\(F.Fygon\)
\(G.Graph\)
\(H.Hash\ Code\ Hacker\)
每次从后一位向前进一位，后一位\(y\)变成\(Z\)，前一位的\(x\)变成\(y\)，重复操作即可
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1111;
char s[MAXN];
int k;
int main(){
#ifdef ONLINE_JUDGE
freopen("hash.in","r",stdin);
#endif
cin >> k;
#ifdef ONLINE_JUDGE
freopen("hash.out","w",stdout);
#endif
for(int i = 1; i <= 999; i++) s[i] = 'x';
s[1000] = 'y';
cout << (s+1) << endl;
for(int i = 1; i < k; i++){
s[1000+1-i] = 'Z';
s[1000-i] = 'y';
cout << (s+1) << endl;
}
return 0;
}\(I.Insider's\ Information\)
\(J.Journey\ to\ the\ "The World's Start"\)
\(K.Kingdom\ Trip\)
\(L.Lucky\ Chances\)
签到，暴力
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 111;
const int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
int n,m,A[MAXN][MAXN];
int main(){
#ifdef ONLINE_JUDGE
freopen("lucky.in","r",stdin);
#endif
____();
cin >> n >> m;
for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> A[i][j];
#ifdef ONLINE_JUDGE
freopen("lucky.out","w",stdout);
#endif
int res = 0;
for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++){
for(int d = 0; d < 4; d++){
bool ok = true;
for(int x = i+dir[d][0], y = j + dir[d][1]; x>0&&x<=n&&y>0&&y<=m; x+=dir[d][0],y+=dir[d][1]) if(A[x][y]>=A[i][j]) ok = false;
res+=ok;
}
}
cout << res << endl;
return 0;
}标签：12,Northern,Contest,int,len,MAXN,freopen,1024000000,tile 来源： https://www.cnblogs.com/kikokiko/p/12243876.html