题解报告（CDUT暑期集训——第二场）
题解报告（CDUT暑期集训——第二场）
D - Game
HDU - 6312
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思路：水题 Alice一直是必胜态
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AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<vector>
#include<stack>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll Pow(ll a, ll b, ll c){
ll ans = 1;
a %= c;
while (b){
if (b & 1) ans = (ans * a) % c;
a = (a * a) % c;
b >>= 1;
}
return (ans % c);
}
bool cmp(const int &a, const int &b){
return a < b;
}
ll gcd(ll x, ll y){
return x % y == 0 ? y : gcd(y, x % y);
}
ll lcm(ll x, ll y){
return x / gcd(x, y) * y;
}
int n;
int main(){
while (scanf("%d", &n) != EOF){
printf("Yes\n");
}
return 0;
}
E - Hack It
HDU - 6313
-
思路：构造矩阵 可构造一个n * n的每行1的个数为\(\sqrt{n}\) 则总的1的个数为n * \(\sqrt{n}\) 题目要求1的总个数不少于85000 n * \(\sqrt{n}\) >= 85000 \(\Rightarrow\) n >= 1933 对n * n的矩阵分块成n个\(\sqrt{n}\) * \(\sqrt{n}\)的矩阵 令p = \(\sqrt{n}\) 从n = \(2 ^ 2\), \(3 ^ 2\), ... （在草稿纸上瞎构造矩阵画出来 发现每一个p * p的小矩阵构成了一个模p循环加群 但是当p为合数时 一定会有重复的 即不满足题意 所有p为素数 由n >= 1933可得p = 47 然后按着递推公式构造矩阵就行了
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AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<vector>
#include<stack>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int N = 5010;
int g[N][N];
int main(){
int p = 47, n = 2000;
int gx, gy;
for (int i = 0; i < p; i ++ ){
for (int j = 0; j < p; j ++ ){
for (int k = 0; k < p; k ++ ){
gx = i * p + j;
gy = k * p + (j * k + i) % p;
g[gx][gy] = 1;
}
}
}
printf("%d\n", n);
for (int i = 0; i < n; i ++ ){
for (int j = 0; j < n; j ++ )
printf("%d", g[i][j]);
printf("\n");
}
return 0;
}
G - Naive Operations
HDU - 6315
-
思路：去洛谷上看的线段树然后套的板子 照着板子改 维护区间内a的最大值和b的最小值就行了 注意区间内a的最大值小于b的最小值时跳出（因为这种情况下 ∀i ∈( l, r) ai / bi = 0 对结果无影响
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AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<vector>
#include<stack>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int N = 100010;
int n, m, l, r;
char op[10];
int b[N], sum[N << 2], tag[N << 2], maxa[N << 2], minb[N << 2];
inline int lson(int x){
return x << 1;
}
inline int rson(int x){
return x << 1 | 1;
}
inline void push_up(int p){
sum[p] = sum[lson(p)] + sum[rson(p)];
maxa[p] = max(maxa[lson(p)], maxa[rson(p)]);
minb[p] = min(minb[lson(p)], minb[rson(p)]);
}
inline void push_down(int p){
if (tag[p]){
maxa[lson(p)] += tag[p];
maxa[rson(p)] += tag[p];
tag[lson(p)] += tag[p];
tag[rson(p)] += tag[p];
tag[p] = 0;
}
}
void build(int p, int l, int r){
tag[p] = 0;
if (l == r){
sum[p] = 0;
maxa[p] = 0;
minb[p] = b[l];
return ;
}
int mid = (l + r) >> 1;
build(lson(p), l, mid);
build(rson(p), mid + 1, r);
push_up(p);
}
inline void update(int L, int R, int l, int r, int p){
if (L <= l && R >= r){
maxa[p] ++;
if (maxa[p] < minb[p]){
tag[p] ++;
return ;
}
if (l == r && maxa[p] >= minb[p]){
sum[p] ++;
minb[p] += b[l];
return ;
}
}
push_down(p);
int mid = (l + r) >> 1;
if (L <= mid) update(L, R, l, mid, lson(p));
if (R > mid) update(L, R, mid + 1, r, rson(p));
push_up(p);
}
int query(int L, int R, int l, int r, int p){
int ans = 0;
if (L <= l && R >= r)
return sum[p];
int mid = (l + r) >> 1;
push_down(p);
if (L <= mid) ans += query(L, R, l, mid, lson(p));
if (R > mid) ans += query(L, R, mid + 1, r, rson(p));
return ans;
}
int main(){
while (scanf("%d%d", &n, &m) != EOF){
for (int i = 1; i <= n; i ++ ) scanf("%d", &b[i]);
build(1, 1, n);
while (m -- ){
scanf("%s%d%d", op, &l, &r);
switch(op[0]){
case 'a':
update(l, r, 1, n, 1);
break;
case 'q':
printf("%d\n", query(l, r, 1, n, 1));
}
}
}
return 0;
}
J - Swaps and Inversions
HDU - 6318
-
思路：就是求逆序数cnt 然后cnt * min(x, y)就行了（最开始写的merge一直TLE 把l写成了1（该打 然后重写了一遍merge竟然过了
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AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<vector>
#include<stack>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int N = 100010;
int a[N], b[N];
ll n, x, y, cnt;
void merge_(ll l, ll r){
if (r - l < 1) return ;
ll mid = (l + r) >> 1;
merge_(l, mid);
merge_(mid + 1, r);
ll i = l, j = mid + 1;
ll pos = 0;
/* for (int k = 1; k <= r; k ++ ){
if (j > r || i <= mid && a[i] < a[j]) b[k] = a[i ++];
else{
b[k] = a[j ++];
cnt += mid - i + 1;
}
}
for (int k = l; k <= r; k ++ ) a[k] = b[k];*/
while (i <= mid && j <= r){
if (a[i] > a[j]){
b[++ pos] = a[j ++];
cnt += mid - i + 1;
}
else b[++ pos] = a[i ++];
}
while (i <= mid) b[++ pos] = a[i ++];
while (j <= r) b[++ pos] = a[j ++];
pos = 0;
for (ll k = l; k <= r; k ++) a[k] = b[++ pos];
}
int main(){
while (scanf("%lld%lld%lld", &n, &x, &y) != EOF){
cnt = 0;
for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
merge_(1, n);
printf("%lld\n", cnt * min(x, y));
}
return 0;
}
