20201103gryz模拟赛解题报告
写在前面
昨天忘写了来补上
T1位运算乱搞一会没搞出来，
打完T4floyd暴力分之后发现T2树状数组可以骗点分
打完T3暴力手模了一遍样例之后发现T3就是个线段树板子
最后就非常愉快的拿到175pts，rank3
T1：U139249 位运算之谜
https://i.cnblogs.com/preference
很显然我并不知道这个公式
因为a&b表示的是都是1的位数，a^b表示的是只有一个是1的位数，
如果都是1，a+b后会进位，否则保留，
所以可推的如上的式子：$a+b=((a&b)<<1)+(a^b) $；
由题意已知\(a&b=y\)，\(a+b=x\)
故\(a^b=x-2 \times y\);
因为这个数取的是不同位，\(y\)取的是相同位，所以他俩\(&\)起来一定为\(0\)
不为 \(0\) 为不合法
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
using namespace std;
long long T, x, y;
long long read(){
long long w = 1, s = 0;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
return s * w;
}
int main()
{
T = read();
while(T--){
x = read(), y = read();
if((x - 2 * y) < 0 || ((x - 2 * y) & y)) cout<<-1<<endl;
else cout<<(x - y - y)<<endl;
}
return 0;
}
T2：U139245 游戏
看到数据范围比较小，预处理一个二维前缀和
暴力枚举每个节点i，j，二分找出最长边长即可
二维树状数组的话还是太慢了
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
using namespace std;
int n, m, K;
int sum[310][310][30];
int a[310][310][30];
int read(){
int w = 1, s = 0;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
return s * w;
}
bool check(int i, int j, int k, int l){
for(int a = 1; a <= 26; ++a){
if(sum[k][l][a] + sum[i - 1][j - 1][a] - sum[i - 1][l][a] - sum[k][j - 1][a] > K) return false;
}
return true;
}
int main()
{
n = read(), m = read(), K = read();
char x;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= m; ++j){
cin>>x;
a[i][j][x - 'a' + 1]++;
for(int k = 1; k <= 26; ++k){
sum[i][j][k] = sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k] + a[i][j][k];
}
}
}
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= m; ++j){
int l = 0, r = min(n - i, m - j), ans = 0;
while(l <= r){
int mid = (l + r) >> 1;
if(check(i, j, i + mid, j + mid)){ ans = mid; l = mid + 1; }
else{ r = mid - 1; }
}
printf("%d ", ans + 1);
}
puts("");
}
return 0;
}
T3：U139247 或和异或
手模样例发现每两个相邻的元素在每次操作后会合并成一个
可以用线段树来维护一下，最后输出树根的答案即可
修改操作对应线段树的单点修改
上传的时候注意判断是or还是xor
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
=============================Kersen AK IOI !!!=====================================================
*/
#include<iostream>
#include<cstdio>
#include<cmath>
#define lson i << 1
#define rson i << 1 | 1
using namespace std;
const int MAXN = 131100;
struct Tree{
long long dep;
long long sum;
}tree[MAXN << 2];
long long n, Q;
bool flag = 0;
long long a[131080];
int ecm[20] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072};
//long long b[131080];
//int wzd[30];
//long long kersen[30];
long long read(){
long long s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
return s * w;
}
void push_up(int i){
if(!flag){//如果n是奇数，树是n + 1层，上传的时候奇数深度进行|操作，偶数深度进行^操作
if(tree[i].dep % 2){tree[i].sum = (tree[lson].sum | tree[rson].sum);}
else{tree[i].sum = (tree[lson].sum ^ tree[rson].sum);}
}
else{//如果n是偶数，树是n + 1层，上传的时候偶数深度进行|操作，奇数深度进行^操作，与上面相反
if(tree[i].dep % 2){tree[i].sum = (tree[lson].sum ^ tree[rson].sum);}
else{tree[i].sum = (tree[lson].sum | tree[rson].sum);}
}
}
void build(long long i, long long l, long long r, long long dep){
tree[i].dep = dep;
if(l == r) {
tree[i].sum = a[l];
return ;
}
int mid = (l + r) >> 1;
build(lson, l, mid, dep + 1), build(rson, mid + 1, r, dep + 1);
push_up(i);
return ;
}
void add(long long i ,long long l, long long r, long long x, long long k){
if(l == x && r == x) {
tree[i].sum = k;
return ;
}
int mid = (l + r) >> 1;
if(x <= mid) add(lson, l, mid, x, k);
else add(rson, mid + 1, r, x, k);
push_up(i);
return ;
}
int main()
{
// freopen("xor.in","r",stdin);
// freopen("xor.out","w",stdout);
n = read(), Q = read();
if(n%2 == 0) flag = 1;//如果n是偶数，标记一下
for(long long i = 1; i <= ecm[n]; ++i) a[i] = read();
build(1, 1, ecm[n], 1);
for(long long i = 1, x, k; i <= Q; ++i){
x = read(), k = read();
add(1, 1, ecm[n], x, k);
cout<<tree[1].sum<<endl;
}
return 0;
}
T4：U139253 链接
因为边的长度随编号的增加而增加，且前i条边的和一定不会超过第i+1条边
根据这个性质，在读入边的时候直接建出最小生成树，就求完了各个点的最短路
考虑u，v这两个节点 ,假设我们已经知道了u，那么我们可以通过如下式子推得v
\[ans_{v} = ans_{u} - siz_{v} \times e_{i}.w + (point - siz_{v}) \times e_{i}.w
\]
最后求一下和就是答案
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#define int long long
using namespace std;
const int MAXN = 1e5+5;
const int MAXM = 2e5+5;
const int mod = 1000000007;
struct edge{
int to, w, nxt;
}e[MAXM];
int head[MAXN], num_edge;
int n, m, point;
int dis[MAXN], fath[MAXN], f[MAXN], siz[MAXN];
bool type[MAXN];
int read(){
int w = 1, s = 0;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = s* 10 + ch - '0', ch = getchar();
return s * w;
}
//bool read1(){
// bool s = 0;
// char ch = getchar();
// while(ch >= '0' && ch <= '9') s = s* 10 + ch - '0', ch = getchar();
// return s;
//}
void add(int from, int to, int w){
e[++num_edge].to = to;
e[num_edge].w = w;
e[num_edge].nxt = head[from];
head[from] = num_edge;
}
int find(int x){return fath[x] == x ? x : fath[x] = find(fath[x]); }
void dfs(int x, int fa){
for(int i = head[x]; i; i = e[i].nxt){
int v = e[i].to;
if(v == fa) continue;
dis[v] = (dis[x] + e[i].w) % mod;
dfs(v, x), siz[x] += siz[v];
}
}
void dfs2(int x, int fa){
for(int i = head[x]; i; i = e[i].nxt){
int v = e[i].to;
if(v == fa) continue;
int jia = ((point - siz[v]) % mod + mod) % mod;
int jian = siz[v] * e[i].w % mod;
jia = (jia * e[i].w) % mod;
f[v] = ((f[x] + jia - jian) % mod + mod) % mod;
dfs2(v, x);
}
}
signed main()
{
n = read(), m = read();
bool flag = 0, se = 0;
se = read();
if(se) flag = 1, type[1] = se ^ 1;
else type[1] = 0;
for(int i = 2; i <= n; ++i){
type[i] = read();
// cin>>type[i];
if(flag) type[i] = (type[i] ^ 1);
siz[i] = type[i];//顺便处理在以i为根的子树中，与1节点不同的点的个数
}
for(int i = 1; i <= n; ++i) fath[i] = i;
int x = 2, cnt = 0;
for(int i = 1, u, v; i <= m; ++i){
u = read(), v = read();
int uf = find(u), vf = find(v);
if(uf != vf){
fath[uf] = vf;
add(u, v, x), add(v, u, x);
//cnt++;
}
// if(++cnt == n - 1) break;
x = (x * 2) % mod;
}
dfs(1, 1), point = siz[1];
for(int i = 1; i <= n; ++i){
if(type[i]) f[1] = (f[1] + dis[i]) % mod;//先暴力把1与其他点的权值和加起来
}
dfs2(1, 1);
int ans = 0;
for(int i = 1; i <= n; ++i){
if(!type[i]) ans = (ans + f[i]) % mod;
}
printf("%d", ans);
return 0;
}
