练习

今天没事把去年浙江省赛题做了做，2个半小时 是个分界线，出8道后就一直划水了。菜鸡一个

A

水题

B

题意：找有多少对(a, b)使得a Xor b > max(a, b)

考虑高位的1，扫一遍就好了。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<map>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
const int MAXN = 1e5 + 10;
typedef long long LL;
int a[MAXN];
int bit[40], len, num[40];
void get(int n) {
len = 0;
while(n) {
bit[len++] = n % 2;
n /= 2;
}
}
int main()
{
int t; scanf("%d", &t);
while(t--) {
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
CLR(num, 0); sort(a+1, a+n+1);
LL ans = 0;
for(int i = 1; i <= n; i++) {
get(a[i]);
for(int j = 0; j < len; j++) {
if(bit[j] == 0) {
ans += num[j];
}
}
num[len-1]++;
}
printf("%lld\n", ans);
}
return 0;
}

C

懵懂

D

题意：求序列所有连续子序列的(去重后)元素和。

思路：考虑每一个元素的贡献，a[i]的贡献为(i - s) * (n - i + 1) * a[i]，其中s-1是上一个a[i]元素出现的位置，扫描一遍就OK了。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<map>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 10;
int vis[MAXN], a[MAXN];
int main()
{
int t; scanf("%d", &t);
while(t--) {
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
vis[a[i]] = 0;
}
LL ans = 0;
for(int i = 1; i <= n; i++) {
if(vis[a[i]] > 0) {
ans += 1LL * (i - vis[a[i]]) * (n - i + 1) * a[i];
}
else {
ans += 1LL * i * (n - i + 1) * a[i];
}
vis[a[i]] = i;
}
printf("%lld\n", ans);
}
return 0;
}

E、F神题

G、H水题

I过不了

J模拟

K模拟

并没有什么坑点

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<map>
#define CLR(a, b) memset(a, (b), sizeof(a))
#define eps 1e-5
using namespace std;
typedef long long LL;
bool vis[110][110][20];
int Map[110][20];
struct Node {
double score;
int id;
};
Node team[110];
bool cmp(Node a, Node b) {
return a.score > b.score;
}
bool cmp1(Node a, Node b) {
return a.id < b.id;
}
int main()
{
int t; scanf("%d", &t);
while(t--) {
int n, p, s, q; scanf("%d%d%d%d", &n, &p, &s, &q);
for(int i = 1; i <= n; i++) {
team[i].score = s; team[i].id = i;
for(int j = 1; j <= p; j++) {
Map[i][j] = 1;
}
}
while(q--) {
int N; scanf("%d", &N); CLR(vis, false);
int rec[110][20][110], num[110][20];
bool attacked[110][20];
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= p; j++) {
attacked[i][j] = false;
num[i][j] = 0;
}
}

while(N--) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
if(vis[x][y][z]) continue;

attacked[y][z] = true;
rec[y][z][num[y][z]] = x;
num[y][z]++;

vis[x][y][z] = true;
Map[y][z] = 0;
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= p; j++) {
if(attacked[i][j]) {
team[i].score -= n - 1;
double total = n - 1;
double add = total / num[i][j];
for(int k = 0; k < num[i][j]; k++) {
team[rec[i][j][k]].score += add;
}
}
}
}
//            for(int i = 1; i <= n; i++) {
//                printf("%.5lf %d\n", team[i].score, team[i].id);
//            }
for(int i = 1; i <= p; i++) {
for(int j = 1; j <= n; j++) {
int v; scanf("%d", &v);
Map[j][i] = v;
}
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= p; j++) {
if(Map[i][j] == 0) {
team[i].score -= n - 1;
int cnt = 0;
for(int k = 1; k <= n; k++) {
if(Map[k][j] && k != i) {
cnt++;
}
}
double add = (n - 1) * 1.0 / cnt;
for(int k = 1; k <= n; k++) {
if(Map[k][j] && k != i) {
team[k].score += add;
}
}
}
}
}
sort(team + 1, team + n + 1, cmp);
int M; scanf("%d", &M);
while(M--) {
int index; scanf("%d", &index);
int Rank = 1;
for(int i = 1; i <= n; i++) {
if(i == 1 || (team[i-1].score - team[i].score) > eps) {
Rank = i;
}
if(team[i].id == index) {
if(i == 1 || (team[i-1].score - team[i].score) < eps) {
printf("%.5lf %d\n", team[i].score, Rank);
}
else {
printf("%.5lf %d\n", team[i].score, i);
}
break;
}
}
}
sort(team + 1, team + n + 1, cmp1);
}
}
return 0;
}

L水题