codeforces-1136 (div2)
A.读到第i章,就有N - i + 1章还没读。
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#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 = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}
while (c >= ''&&c <= ''){x = x * + c - '';c = getchar();}return x*f;}
const double eps = 1e-;
const int maxn = ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,K;
PII a[maxn];
int main(){
Sca(N);
for(int i = ; i <= N ; i ++){
Sca2(a[i].fi,a[i].se);
}
K = read();
for(int i = ; i <= N ; i ++){
if(a[i].fi <= K && K <= a[i].se){
Pri(N - i + );
break;
}
}
return ;
}
B.很显然,先把起始点的石头扔到一边,然后向离端点距离短的一边走,走到端点再想另一端走,途中一边取硬币,石头全扔到起始点。
推出式子显然
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#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 = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}
while (c >= ''&&c <= ''){x = x * + c - '';c = getchar();}return x*f;}
const double eps = 1e-;
const int maxn = ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,K;
PII a[maxn];
int main(){
Sca2(N,K);
Pri( + N + * min(N - K,K - ) + max(N - K,K - ) + N - );
return ;
}
C.发现对2 * 2的矩阵操作就是把斜相邻的两个数交换,其余N * M的矩阵操作都可以通过交换两个斜对角实现,所以交换只能交换左上右下这样一个斜对角。
那么判断两个矩阵的每个斜对角是否是相同的元素即可
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#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 = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}
while (c >= ''&&c <= ''){x = x * + c - '';c = getchar();}return x*f;}
const double eps = 1e-;
const int maxn = ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,K;
int A[maxn][maxn],B[maxn][maxn];
vector<int>a[maxn * ],b[maxn * ];
int main(){
Sca2(N,M);
For(i,,N) For(j,,M) Sca(A[i][j]);
For(i,,N) For(j,,M) Sca(B[i][j]);
For(i,,N) For(j,,M){
a[i + j].pb(A[i][j]);
b[i + j].pb(B[i][j]);
}
for(int i = ; i <= N + M ; i ++){
sort(a[i].begin(),a[i].end());
sort(b[i].begin(),b[i].end());
if(a[i] != b[i]){
puts("NO");
return ;
}
}
puts("YES");
return ;
}
D.前面的u点可以和后面的v点交换称为u可以吃掉v,点i最终可以排到nastya的后面需要满足在他后面的点是自主排到nastya后面或者它可以被i吃掉。
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#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 = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}
while (c >= ''&&c <= ''){x = x * + c - '';c = getchar();}return x*f;}
const double eps = 1e-;
const int maxn = 3e5 + ;
const int maxm = 5e5 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,K;
int a[maxn];
int pos[maxn];
int pre[maxn];
struct Edge{
int to,next;
}edge[maxm * ];
int head[maxn],tot;
void init(){
for(int i = ; i <= N ; i ++) head[i] = -;
tot = ;
}
void add(int u,int v){
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
int main(){
Sca2(N,M); init();
for(int i = ; i <= N ; i ++){
a[i] = read();
pos[a[i]] = i;
}
for(int i = ; i <= M ; i ++){
PII e; e.fi = read(); e.se = read();
if(pos[e.fi] < pos[e.se]){
pre[pos[e.fi]]++;
add(pos[e.se],pos[e.fi]);
}
}
int ans = ,now = ;
for(int i = N - ; i >= ; i --){
if(pre[i] == now){
ans++;
for(int j = head[i]; ~j ; j = edge[j].next) pre[edge[j].to]--;
}else{
now++;
}
}
Pri(ans);
return ;
}
D
E.对k求一个前缀和pre,对于u点,属于[u,N]之间的所有点i的最小值都会满足 a[x] >= a[u] + pre[i] - pre[u]
对于i来说,pre[i]是固定的,动态更新的是a[u] - pre[u],所以将pre[i]分离出来,线段树维护每个点的a[u] - pre[u],维护区间和以及最大最小值。
求区间和的时候就是ppre[r] - ppre[l - 1] + 线段树维护的l到r之间的和。
坑点:lazy标记不能打成0,需要设为-INF,因为a[u] - pre[u]可能是负数
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#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;
LL read(){LL x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}
while (c >= ''&&c <= ''){x = x * + c - '';c = getchar();}return x*f;}
const double eps = 1e-;
const int maxn = 1e5 + ;
const LL INF = 1e18;
const int mod = 1e9 + ;
int N,M,K;
struct Tree{
int l,r;
LL sum,MIN,MAX,lazy;
}tree[maxn << ];
LL a[maxn],pre[maxn],ppre[maxn];
void Pushup(int t){
tree[t].sum = tree[t << ].sum + tree[t << | ].sum;
tree[t].MIN = min(tree[t << ].MIN,tree[t << | ].MIN);
tree[t].MAX = max(tree[t << ].MAX,tree[t << | ].MAX);
}
void Build(int t,int l,int r){
tree[t].l = l; tree[t].r = r;
tree[t].lazy = -INF;
if(l == r){
tree[t].MAX = tree[t].sum = tree[t].MIN = a[l] - pre[l];
return;
}
int m = l + r >> ;
Build(t << ,l,m); Build(t << | ,m + ,r);
Pushup(t);
}
void change(int t,LL p){
tree[t].lazy = max(p,tree[t].lazy);
tree[t].sum = (tree[t].r - tree[t].l + ) * p;
tree[t].MAX = tree[t].MIN = p;
}
void Pushdown(int t){
if(tree[t].lazy != -INF){
change(t << ,tree[t].lazy); change(t << | ,tree[t].lazy);
tree[t].lazy = -INF;
}
}
void update(int t,int l,int r,LL p){
if(tree[t].MIN >= p) return;
if(l <= tree[t].l && tree[t].r <= r && tree[t].MAX <= p){
change(t,p);
return;
}
Pushdown(t);
int m = (tree[t].r + tree[t].l) >> ;
if(r <= m) update(t << ,l,r,p);
else if(l > m) update(t << | ,l,r,p);
else{
update(t << ,l,m,p);
update(t << | ,m + ,r,p);
}
Pushup(t);
}
LL query(int t,int l,int r){
if(l <= tree[t].l && tree[t].r <= r) return tree[t].sum;
Pushdown(t);
int m = (tree[t].l + tree[t].r) >> ;
if(r <= m) return query(t << ,l,r);
else if(l > m) return query(t << | ,l,r);
else return query(t << ,l,m) + query(t << | ,m + ,r);
}
int main(){
Sca(N);
for(int i = ; i <= N ; i ++) a[i] = read();
for(int i = ; i <= N; i ++){
pre[i] = read() + pre[i - ];
ppre[i] = pre[i] + ppre[i - ];
}
Build(,,N);
int Q; Sca(Q);
while(Q--){
char op[]; int l,r;
scanf("%s%d%d",&op,&l,&r);
if(op[] == '+'){
LL x = query(,l,l) + r + pre[l];
update(,l,N,x - pre[l]);
}else{
Prl(query(,l,r) + ppre[r] - ppre[l - ]);
}
}
return ;
}
E
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