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1001
00:23:46 solved by hl
签到 枚举一下位就行了
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();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;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
LL A,B;
int main(){
int T = read();
while(T--){
scanf("%lld%lld",&A,&B);
LL c = 0;
for(int i = 0; i <= 32; i ++){
LL t = (1ll << i);
int a = 0,b = 0;
if(A & t) a = 1;
if(B & t) b = 1;
if(a && b) c |= t;
}
if(!c) c++;
Prl(c);
}
return 0;
}
1001
1002
2:40:10(-1) solved by hl,ysw
对序列维护一个权值线段树，维护的东西是区间最大值，然后对于每一个查询二分查找答案，如果在起始点在答案有东西的下标大于终点，就代表答案可以更小
时间复杂度n(logn)²,事实上由于权值线段树的特性，寻找大于R的最小值可以直接在线段树上操作，简化了二分这个操作，时间复杂度nlogn
比赛的时候并没有想到，冲了读入挂趁评测机不注意过去了
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1;
inline bool isdigit(const char& n) {return (n>=‘0‘&&n<=‘9‘);}
inline void rnext(){if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);}
template <class T> inline bool in(T &ans){
#ifdef VSCode
ans=0;T f=1;register char c;
do{c=getchar();if (‘-‘==c)f=-1;}while(!isdigit(c)&&c!=EOF);
if(c==EOF)return false;do{ans=(ans<<1)+(ans<<3)+c-48;
c=getchar();}while(isdigit(c)&&c!=EOF);ans*=f;return true;
#endif
#ifndef VSCode
ans =0;T f=1;if(ps==pe)return false;do{rnext();if(‘-‘==*ps)f=-1;}
while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(ans<<1)+(ans<<3)+*ps-48;
rnext();}while(isdigit(*ps)&&ps!=pe);ans*=f;return true;
#endif
}const int MAXOUT=10000;
char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
inline void write(){fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;}
inline void out_char(char c){*(pout++)=c;if(pout==pend)write();}
inline void out_str(char *s){while(*s){*(pout++)=*(s++);if(pout==pend)write();}}
template <class T>inline void out_int(T x) {if(!x){out_char(‘0‘);return;}
if(x<0)x=-x,out_char(‘-‘);int len=0;while(x){outtmp[len++]=x%10+48;x/=10;}outtmp[len]=0;
for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp);}
template<typename T, typename... T2>
inline int in(T& value, T2&... value2) { in(value); return in(value2...); }
int read(){int x = 0,f = 1;char c = getchar();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;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
int a[maxn],id[maxn];
struct Tree{
int l,r,Max;
}tree[maxn << 2];
inline void Pushup(int t){
tree[t].Max = max(tree[t << 1].Max,tree[t << 1 | 1].Max);
}
inline void Build(int t,int l,int r){
tree[t].l = l; tree[t].r = r;
if(l == r){
tree[t].Max = id[l];
return;
}
int m = l + r >> 1;
Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
Pushup(t);
}
inline int query(int t,int l,int r){
if(l <= tree[t].l && tree[t].r <= r){
return tree[t].Max;
}
int m = (tree[t].l + tree[t].r) >> 1;
if(r <= m) return query(t << 1,l,r);
else if(l > m) return query(t << 1 | 1,l,r);
return max(query(t << 1,l,m),query(t << 1 | 1,m + 1,r));
}
inline void update(int t,int p){
if(tree[t].l == tree[t].r){
tree[t].Max = INF;
return;
}
int m = (tree[t].l + tree[t].r) >> 1;
if(p <= m) update(t << 1,p);
else update(t << 1 | 1,p);
Pushup(t);
}
inline int solve(int k,int R){
int l = k,r = N + 1;
int ans = N + 1;
while(l <= r){
int m = l + r >> 1;
if(query(1,k,m) > R){
ans = m;
r = m - 1;
}else{
l = m + 1;
}
}
return ans;
}
int main(){
int T;
in(T);
while(T--){
in(N); in(M);
for(int i = 1; i <= N ; i ++) in(a[i]),id[a[i]] = i;
id[N + 1] = INF;
Build(1,1,N + 1);
int ans = 0;
while(M--){
int op; in(op);
if(op == 1){
int t; in(t);
t ^= ans;
update(1,a[t]);
}else{
int r,k; in(r); in(k);
r ^= ans; k ^= ans;
// cout << k << " " << r << endl;
Pri(ans = solve(k,r));
}
}
}
return 0;
}
1002
1003
1:31:38 (-1) solved by hl
主席树的tot没有初始化 RE一发
先走一个后缀数组，发现只要在给定起始位置上的Height数组连续长度超过给定子串长度len中的一段
用ST表 + 二分可以log时间求出这一段的左右端点，然后就是一个静态区间K大，主席树即可
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();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;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
char str[maxn];
int sa[maxn],rak[maxn],tex[maxn],tp[maxn],Height[maxn];
void GetHeight() {
int j, k = 0;
for(int i = 1; i <= N; i++){
if(k) k--;
int j = sa[rak[i] - 1];
while(str[i + k] == str[j + k]) k++;
Height[rak[i]] = k;
}
}
void Qsort(){
for(int i = 0; i <= M ; i ++) tex[i] = 0;
for(int i = 1; i <= N ; i ++) tex[rak[i]]++;
for(int i = 1; i <= M ; i ++) tex[i] += tex[i - 1];
for(int i = N; i >= 1 ; i --) sa[tex[rak[tp[i]]]--] = tp[i];
}
void SA(){
for(int i = 1; i <= N ; i ++) rak[i] = str[i] - ‘a‘ + 1,tp[i] = i;
Qsort();
for(int w = 1,p = 0; p < N; w <<= 1,M = p){
p = 0;
for(int i = 1; i <= w; i ++) tp[++p] = N - w + i;
for(int i = 1; i <= N ; i ++) if(sa[i] > w) tp[++p] = sa[i] - w;
Qsort();
swap(tp,rak);
rak[sa[1]] = p = 1;
for(int i = 2; i <= N ; i ++){
rak[sa[i]] = (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])?p:++p;
}
}
}
int T[maxn];
struct Tree{
int lt,rt,sum;
void init(){
lt = rt = sum = 0;
}
}tree[maxn * 32];
int tot;
void newnode(int &t){
t = ++tot;
tree[t].init();
}
void Build(int &t,int l,int r){
newnode(t);
if(l == r) return;
int m = l + r >> 1;
Build(tree[t].lt,l,m); Build(tree[t].rt,m + 1,r);
}
void update(int &t,int pre,int l,int r,int p){
newnode(t);
tree[t] = tree[pre];
tree[t].sum++;
if(l == r) return;
int m = l + r >> 1;
if(p <= m) update(tree[t].lt,tree[pre].lt,l,m,p);
else update(tree[t].rt,tree[pre].rt,m + 1,r,p);
}
int query(int L,int R,int l,int r,int k){
if(k > tree[R].sum - tree[L].sum) return -1;
if(l >= r) return l;
int num = tree[tree[R].lt].sum - tree[tree[L].lt].sum;
int m = l + r >> 1;
if(num >= k) return query(tree[L].lt,tree[R].lt,l,m,k);
else return query(tree[L].rt,tree[R].rt,m + 1,r,k - num);
}
const int SP = 20;
int MIN[maxn][SP];
int mm[maxn];
void initRMQ(int n,int b[]){
for(int i = 1; i <= n ; i ++){
for(int j = 0; j < SP; j ++){
MIN[i][j] = INF;
}
}
mm[0] = -1;
for(int i = 1; i <= n ; i ++){
mm[i] = ((i & (i - 1)) == 0)?mm[i - 1] + 1:mm[i - 1];
MIN[i][0] = b[i];
}
for(int j = 1; j <= mm[n]; j ++){
for(int i = 1; i + (1 << j) - 1 <= n ; i ++){
MIN[i][j] = min(MIN[i][j - 1],MIN[i + (1 << (j - 1))][j - 1]);
}
}
}
int rmq(int x,int y){
if(x > y) return -1;
int k = mm[y - x + 1];
return min(MIN[x][k],MIN[y - (1 << k) + 1][k]);
}
int main(){
int TTT = read();
while(TTT--){
Sca2(N,K); tot = 0;
scanf("%s",str + 1);
M = 122;
SA(); GetHeight();
Build(T[0],1,N);
for(int i = 1; i <= N ; i ++) update(T[i],T[i - 1],1,N,sa[i]);
initRMQ(N,Height);
while(K--){
int l,r,k; Sca3(l,r,k);
int len = r - l + 1;
int p = rak[l];
int L = p + 1,R = p;
int ll = 2,rr = p;
while(ll <= rr){
int mid = ll + rr >> 1;
if(rmq(mid,p) >= len){
rr = mid - 1;
L = mid;
}else{
ll = mid + 1;
}
}
L--;
ll = p + 1,rr = N;
while(ll <= rr){
int mid = ll + rr >> 1;
if(rmq(p + 1,mid) >= len){
ll = mid + 1;
R = mid;
}else{
rr = mid - 1;
}
}
Pri(query(T[L - 1],T[R],1,N,k));
}
}
return 0;
}
1003
1004
3:37:37(-2) solved by hl
做法
1.离线所有询问
2.将一个(点，当前距离)的二元组不断扔进堆，每次取出一个二元组然后丢进所有这个点引申出去的二元组
3.直到取出前50000个，所有答案就都出来了
注意点
1.堆内只能容忍最多5w个二元组，不限制会MLE，所以要用multiset弹出最大的
2.菊花图可以轻松卡掉，如果不边排序的话
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1;
inline bool isdigit(const char& n) {return (n>=‘0‘&&n<=‘9‘);}
inline void rnext(){if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);}
template <class T> inline bool in(T &ans){
#ifdef VSCode
ans=0;T f=1;register char c;
do{c=getchar();if (‘-‘==c)f=-1;}while(!isdigit(c)&&c!=EOF);
if(c==EOF)return false;do{ans=(ans<<1)+(ans<<3)+c-48;
c=getchar();}while(isdigit(c)&&c!=EOF);ans*=f;return true;
#endif
#ifndef VSCode
ans =0;T f=1;if(ps==pe)return false;do{rnext();if(‘-‘==*ps)f=-1;}
while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(ans<<1)+(ans<<3)+*ps-48;
rnext();}while(isdigit(*ps)&&ps!=pe);ans*=f;return true;
#endif
}const int MAXOUT=10000;
char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
inline void write(){fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;}
inline void out_char(char c){*(pout++)=c;if(pout==pend)write();}
inline void out_str(char *s){while(*s){*(pout++)=*(s++);if(pout==pend)write();}}
template <class T>inline void out_int(T x) {if(!x){out_char(‘0‘);return;}
if(x<0)x=-x,out_char(‘-‘);int len=0;while(x){outtmp[len++]=x%10+48;x/=10;}outtmp[len]=0;
for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp);}
template<typename T, typename... T2>
inline int in(T& value, T2&... value2) { in(value); return in(value2...); }
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,Q;
struct Edge{
int to;
LL dis;
Edge(){}
Edge(int to,LL dis):to(to),dis(dis){}
};
bool cmp(Edge a,Edge b){
return a.dis < b.dis;
}
vector<Edge>MAP[maxn];
PII query[maxn];
struct node{
int pos;
LL dis;
node(){}
node(int pos,LL dis):pos(pos),dis(dis){}
friend bool operator < (node a,node b){
return a.dis < b.dis;
}
};
LL ans[maxn];
int main(){
int T; in(T);
while(T--){
in(N); in(M); in(Q);
for(int i = 0 ; i <= N; i ++) MAP[i].clear();
for(int i = 1; i <= M ; i ++){
int u,v; LL w;
in(u); in(v); in(w);
MAP[u].push_back(Edge(v,w));
}
for(int i = 1; i <= N ; i ++) sort(MAP[i].begin(),MAP[i].end(),cmp);
for(int i = 1; i <= Q; i ++){
in(query[i].fi);
query[i].se = i;
}
sort(query + 1,query + 1 + Q);
multiset<node>P;
for(int i = 1; i <= N ; i ++){
for(int j = 0 ; j < MAP[i].size(); j ++){
P.insert(node(MAP[i][j].to,MAP[i][j].dis));
// cout << MAP[i][j].to << " " << MAP[i][j].dis << endl;
}
}
multiset<node>::iterator it;
int sum = query[Q].fi;
LL ed = 4e18;
while(P.size() > sum){
it = P.end(); it--;ed = (*it).dis;
P.erase(it);
}
int k = 0;
int cnt = 1;
while(sum--){
it = P.begin();
node u = *it; k++; P.erase(it);
while(cnt <= Q && query[cnt].fi == k) ans[query[cnt].se] = u.dis,cnt++;
for(int i = 0 ; i < MAP[u.pos].size(); i ++){
int v = MAP[u.pos][i].to;
if(MAP[u.pos][i].dis + u.dis >= ed) break;
P.insert(node(v,MAP[u.pos][i].dis + u.dis));
while(P.size() > sum){
it = P.end(); it--;
ed = (*it).dis;
P.erase(it);
}
}
}
for(int i = 1; i <= Q; i ++){
Prl(ans[i]);
}
}
return 0;
}
1004
1006
0:18:13 solved by hl
签到，从后往前依次输出即可
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();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;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
bool vis[maxn];
int a[maxn],b[maxn];
int main(){
while(~Sca2(N,M)){
for(int i = 0; i <= N ; i ++) vis[i] = 0;
for(int i = 1; i <= N ; i ++){
Sca(a[i]);
}
for(int i = 1; i <= M ; i ++){
Sca(b[i]);
}
for(int i = M ; i >= 1; i --){
if(!vis[b[i]]){
printf("%d ",b[i]);
vis[b[i]] = 1;
}
}
for(int i = 1; i <= N ; i ++){
if(!vis[a[i]]){
printf("%d ",a[i]);
vis[a[i]] = 1;
}
}
}
return 0;
}
1006
1007
0:13:46 solved by hl
签到，按照规律弄出一个大的矩阵然后输出即可
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();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;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 2010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
int MAP[maxn][maxn];
int main(){
MAP[1][1] = MAP[1][2] = MAP[2][2] = 1;
MAP[2][1] = 0;
for(int i = 2; i <= 10; i ++){
int s = (1LL << (i - 1));
int t = s;
for(int j = 1; j <= s; j ++){
for(int k = 1; k <= s; k ++){
MAP[j][k + s] = MAP[j][k];
MAP[j + s][k + s] = MAP[j][k];
MAP[j + s][k] = MAP[j][k] ^ 1;
}
}
}
int T = read();
while(T--){
Sca(N);
int s = (1LL << N);
for(int i = 1; i <= s; i ++){
for(int j = 1; j <= s; j ++){
if(MAP[i][j]) printf("C");
else putchar(‘P‘);
}
puts("");
}
}
return 0;
}
1007
1008
1:54:25(-1) solved by zcz
因hl冲了一发贪心而产生的WA
队友写的 我不太明白怎么做
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
long long a[100005];
long long b[100005];
int main()
{
int T;
cin>>T;
while(T--)
{
int n,k;
cin>>n>>k;
long long s1=0;
long long s2=0;
for(int i=0;i<n;i++)
{
scanf("%lld",&a[i]);
s1+=a[i]/k;
s2+=a[i];
b[i]=k-(a[i]%k);
}
sort(b,b+n);
for(int i=0;i<n-(s1+1);i++)
{
s2+=b[i];
}
cout<<s2+k<<endl;
}
return 0;
}
1008
原文地址：https://www.cnblogs.com/Hugh-Locke/p/11408509.html
时间： 2024-11-05 20:37:15