E. Inna and Binary Logic
显然对于一次更新应该一位一位的来，对于第k位的改变，通过找规律可以发现，被更新的数字数量为(k前面有多少个连续的1 + 1) * (k后面有多少个连续的1 + 1) ，找到这个规律时候，直接开线段树或者用set维护连续是1的区间就好了。
这里我是用set写的，毕竟时间给的比较松，STL果然是方便啊。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
#define MP(a,b) make_pair(a,b)
typedef long long LL;
typedef pair<int,int> pii;
const int maxn = 100000 + 5;
int n,m;
int a[maxn];
set<pii> seg[32];
LL add(set<pii> &sg,int pos,int k) {
pii now = MP(pos,n + 1);
set<pii>::iterator it = sg.lower_bound(now);
int cnt_prev = 0,cnt_next = 0,nl = pos,nr = pos;
if(it != sg.begin()) {
it--;
if(it->second == pos - 1) {
cnt_prev = it->second - it->first + 1;
nl = it->first; sg.erase(it);
}
}
it = sg.lower_bound(now);
if(it != sg.end()) {
if(it->first == pos + 1) {
cnt_next = it->second - it->first + 1;
nr = it->second; sg.erase(it);
}
}
sg.insert(MP(nl,nr));
return (1LL << k) * (cnt_prev + 1) * (cnt_next + 1);
}
LL del(set<pii> &sg,int pos,int k) {
pii now = MP(pos,n + 1);
set<pii>::iterator it = sg.upper_bound(now);
it--;
int cnt_prev = pos - it->first,cnt_next = it->second - pos;
if(pos != it->first) sg.insert(MP(it->first,pos - 1));
if(pos != it->second) sg.insert(MP(pos + 1,it->second));
sg.erase(it);
return (1LL << k) * (cnt_prev + 1) * (cnt_next + 1) * -1;
}
LL update(int p,int v) {
LL ret = 0;
for(int i = 0;i < 32;i++) {
int bita = (bool)(a[p] & (1 << i)),bitb = (bool)(v & (1 << i));
if(bita == 0 && bitb == 1) ret += add(seg[i],p,i);
if(bita == 1 && bitb == 0) ret += del(seg[i],p,i);
}
a[p] = v;
return ret;
}
int main() {
scanf("%d%d",&n,&m);
LL sum = 0;
for(int i = 1;i <= n;i++) {
int tmp; scanf("%d",&tmp);
sum += update(i,tmp);
}
for(int i = 1;i <= m;i++) {
int p,v; scanf("%d%d",&p,&v);
sum += update(p,v);
cout << sum << endl;
}
return 0;
}
D. Dima and Bacteria
先开一个并查集然后只把权值为0的边插进去统计同一种类的是否都在一个集合里面，如果不是的话直接No
然后因为同一种类型的细菌相互之间转化肯定是0，因此加入所有的边，只要两边不属于同一种类就更新种类之间的转换关系即可。
快速判断一个点属于哪一类可以先对c数组求前缀和，然后进行二分查找即可。
最后做一遍Folyd直接输出矩阵即可
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 100000 + 5;
const int maxk = 600;
const int INF = INT_MAX / 3;
int fa[maxn],c[maxk];
int n,m,k,mp[maxk][maxk];
int findp(int x) {
return x == fa[x] ? x : fa[x] = findp(fa[x]);
}
inline int getid(int x) {
return lower_bound(c + 1,c + 1 + k,x) - c;
}
void solve() {
for(int p = 1;p <= k;p++) {
for(int i = 1;i <= k;i++) {
for(int j = 1;j <= k;j++) {
mp[i][j] = min(mp[i][j],mp[i][p] + mp[p][j]);
}
}
}
puts("Yes");
for(int i = 1;i <= k;i++) {
for(int j = 1;j <= k;j++) {
if(j > 1) putchar(' ');
if(mp[i][j] >= INF) printf("-1");
else printf("%d",mp[i][j]);
}
puts("");
}
}
int main() {
scanf("%d%d%d",&n,&m,&k);
for(int i = 1;i <= k;i++) scanf("%d",&c[i]);
for(int i = 1;i <= k;i++) c[i] += c[i - 1];
for(int i = 1;i <= k;i++) {
for(int j = 1;j <= k;j++) {
mp[i][j] = INF;
}
mp[i][i] = 0;
}
for(int i = 1;i <= n;i++) fa[i] = i;
int u,v,w,pa,pb,idu,idv;
for(int i = 1;i <= m;i++) {
scanf("%d%d%d",&u,&v,&w);
if(w == 0) {
pa = findp(u); pb = findp(v);
if(pa != pb) fa[pa] = pb;
}
idu = getid(u); idv = getid(v);
mp[idu][idv] = mp[idv][idu] = min(mp[idu][idv],w);
}
bool bad = false;
for(int i = 1;i <= k;i++) {
int nowfa = findp(c[i - 1] + 1);
for(int j = c[i - 1] + 1;j <= c[i];j++) {
if(findp(j) != nowfa) {
bad = true; break;
}
}
if(bad) break;
}
if(bad) puts("No");
else solve();
return 0;
}
C. Inna and Huge Candy Matrix
本质上之后两种变化，一种是转置，一种是水平翻转。知道了之后随便搞一下就好了
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
int n,m,a,b,c,p;
void swap(int &a,int &b) {
int tmp = a; a = b; b = tmp;
}
void hr(int &x,int &y,int &n,int &m) {
y = m - y + 1;
}
void c90(int &x,int &y,int &n,int &m) {
swap(x,y); swap(n,m);
hr(x,y,n,m);
}
void rc90(int &x,int &y,int &n,int &m) {
hr(x,y,n,m);
swap(x,y); swap(n,m);
}
void work(int x,int y,int n,int m) {
for(int i = 0;i < a % 4;i++) c90(x,y,n,m);
for(int i = 0;i < b % 2;i++) hr(x,y,n,m);
for(int i = 0;i < c % 4;i++) rc90(x,y,n,m);
printf("%d %d
",x,y);
}
int main() {
scanf("%d%d%d%d%d%d",&n,&m,&a,&b,&c,&p);
for(int i = 1;i <= p;i++) {
int nx,ny; scanf("%d%d",&nx,&ny);
work(nx,ny,n,m);
}
return 0;
}
B. Inna and New Matrix of Candies
水题，统计一下有几个不同的距离，有负数直接输出不行
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 1005;
set<int> st;
int main() {
int n,m; scanf("%d%d",&n,&m);
bool bad = false;
for(int i = 1;i <= n;i++) {
int ns,ng;
for(int j = 1;j <= m;j++) {
char tmp; scanf(" %c",&tmp);
if(tmp == 'G') ng = j;
if(tmp == 'S') ns = j;
}
if(ns - ng < 0) bad = true;
st.insert(ns - ng);
}
if(bad) puts("-1");
else printf("%d
",(int)st.size());
return 0;
}
A. 随便模拟一下就好
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 15;
char buf[maxn];
char mp[maxn][maxn];
vector<int> aa,ab;
int main() {
int T; scanf("%d",&T);
while(T--) {
scanf("%s",buf + 1);
aa.clear(); ab.clear();
for(int r = 1;r <= 12;r++) if(12 % r == 0) {
int cnt = 1;
for(int i = 1;i <= r;i++) {
for(int j = 1;j <= 12 / r;j++) {
mp[i][j] = buf[cnt++];
}
}
bool ok = false;
for(int j = 1;j <= 12 / r;j++) {
ok = true;
for(int i = 1;i <= r;i++) {
if(mp[i][j] != 'X') ok = false;
}
if(ok) break;
}
if(ok) {
aa.push_back(r); ab.push_back(12 / r);
}
}
printf("%d",(int)aa.size());
for(int i = 0;i < aa.size();i++) printf(" %dx%d",aa[i],ab[i]);
puts("");
}
return 0;
}