A Got Any Grapes?
顺序判断即
比赛的时候忘记写 else 既然 PP 了，然后就 FST
#include <cstdio>
int an,dm,mi;
int gr,pu,bl;
int main(){
scanf("%d%d%d", &an, &dm, &mi);
scanf("%d%d%d", &gr, &pu, &bl);
if(an > gr) {
printf("NO\n");
return 0;
}
else gr -= an;
if(pu + gr < dm){
printf("NO\n");
return 0;
}
else {
if(pu <= dm) {dm -= pu; pu = 0;}
else {pu -= dm; dm = 0;}
if(gr < dm) {
printf("NO\n");
return 0;
}
else gr -= dm;
}
if(gr + pu + bl < mi) printf("NO\n");
else printf("YES\n");
}B Yet Another Array Partitioning Task
反正最后还是要前$ m \times k $ 个数字
直接离散化，然后见当前区间有$ m $ 个就收
#include <cstdio>
#include <algorithm>
const long long N = 2e5 + 1e4;
struct node{
long long now, id;
}b[N];
long long n, m, k, cnt, time;
long long a[N];
bool vis[N];
bool cmp(node a, node b){
return a.now > b.now;
}
int main(){
scanf("%lld%lld%lld", &n, &m, &k);
for(long long i = 1; i <= n; i++){
scanf("%lld", &a[i]);
b[i].now = a[i], b[i].id = i;
}
std::sort(b + 1, b + n + 1, cmp);
long long tmp = m * k;
for(long long i = 1; i <= tmp; i++) vis[ b[i].id ] = true, cnt += b[i].now;
printf("%lld\n",cnt);
cnt = 0;
for(long long i = 1; i <= n; i++){
cnt += vis[i];
if(cnt == m){
time++;
time < k && printf("%lld ",i);
cnt = 0;
}
}
}C. Trailing Loves (or L’oeufs?)
求 $ b $ 进制下 $ n! $ 的末尾的 0 的个数
显然我们要求中间乘出来有多少个$b$
接下来就是分解质因数，然后枚举求最小值即可
#include <cstdio>
#include <cmath>
const long long INF = 1e18 + 1e17;
inline long long Min(long long a, long long b){return a < b? a: b;}
struct node{
long long now, cnt;
}pri[(int)(1e6)];
int pcnt;
long long n, b, tmp, cnt, ans = INF;
inline void get_pri(long long b){
tmp = std::sqrt(b);
for(long long i = 2; i <= tmp; i++){
if(b % i == 0){
pri[ ++pcnt ] = (node){i, 0};
while(b % i == 0){
pri[pcnt].cnt++;
b/=i;
}
}
}
if(b != 1) pri[ ++pcnt ] = (node){b, 1};
}
int main(){
scanf("%lld%lld", &n, &b);
get_pri(b);
for(int i = 1; i <= pcnt; i++){
tmp = 1; cnt = 0;
while(tmp * pri[i].now <= n){
tmp *= pri[i].now;
if(tmp < 0 || tmp % pri[i].now != 0) break;
cnt += n / tmp;
}
ans = Min(ans, cnt/pri[i].cnt);
}
printf("%lld\n", ans);
}D. Flood Fil
有一个非常显然的地方，就是我们每次有两个状态，当前联通部分向左对其或者想右对其
然后我们先把数列中的联通部分预处理出来，然后区间 dp 即可
#include <cstdio>
#include <cstring>
inline int Min(int a, int b){return a < b? a: b;}
const int N = 5100;
int n;
int a[N], l[N], r[N], f[N][N];
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
l[1] = 1;
for(int i = 2; i <= n; i++){
if(a[i] == a[i - 1]) l[i] = l[i - 1];
else l[i] = i;
}
r[n] = n;
for(int i = n - 1; i >= 1; i--){
if(a[i] == a[i + 1]) r[i] = r[i + 1];
else r[i] = i;
}
memset(f, 0x3f, sizeof(f));
for(int i = 1; i<= n;i ++) f[ l[i] ][ r[i] ] = 0;
for(int len = 0; len < n; len++){
for(int i = 1,j; i + len <= n; i++){
j = i + len;
if(i > 1 && j < n && a[i - 1] == a[j + 1])
f[ l[i - 1] ][ r[j + 1] ] = Min(f[ l[i - 1] ][ r[j + 1] ],f[i][j] + 1);
if(i > 1)
f[ l[i - 1] ][j] = Min(f[ l[i - 1] ][j], f[i][j] + 1);
if(j < n)
f[i][ r[j + 1] ] = Min(f[i][ r[j + 1] ], f[i][j] + 1);
}
}
printf("%d", f[1][n]);
}E. Arithmetic Progression
交互题目，60次询问内知道当前乱序等差数列的首项和公差
先二分找最大的，然后random_shuffle随机化询问，求与最大项差的 gcd 即为公差
#include <cstdio>
#include <algorithm>
int gcd(int a, int b){return b? gcd(b ,a%b): a;}
inline int Min(int a, int b){return a < b? a: b;}
inline int Aabs(int a){return a < 0? (0 - a): a;}
const int N = 1e6 + 1e5;
int n, global_tmp, d, max, chance_cnt = 60;
int a[N];
inline bool has_num(int now){
printf("> %d\n",now);
fflush(stdout);
scanf("%d", &global_tmp);
chance_cnt--;
return global_tmp;
}
inline int binary_search_max(int max_limit){
int left = 0, rig = max_limit, mid, res;
while(left <= rig){
mid = (left + rig) >> 1;
if( has_num(mid) ) left = mid + 1;
else rig = mid - 1, res = mid;
}
return res;
}
int main(){
scanf("%d", &n);
max = binary_search_max(1e9);
for(int i = 1; i <= n; i++) a[i] = i;
for(int i = 1; i <= 5; i++) std::random_shuffle(a + 1, a + n + 1);
for(int i = 1; i <= Min(n ,chance_cnt); i++){
printf("? %d\n", a[i]);
fflush(stdout);
scanf("%d", &global_tmp);
if(global_tmp == max) continue;
if(d == 0) d = Aabs(max-global_tmp);
d = gcd(d , Aabs(max-global_tmp));
}
fflush(stdout);
printf("! %d %d\n", max - (n - 1) * d, d);
}F. Please, another Queries on Array?
这个题目首先得知道$\varphi(n)$的求法
然后就是乘积线段树和或线段树维护一下
就没有然后了
#include <cstdio>
const int N = 4e5 + 1e4;
const long long mod = 1e9 + 7;
int n, q, x, y, z, pcnt;
int a[N];
long long p[310], pri[N], inv[310];
char op[10];
inline long long ksm(long long a, long long p){
long long res = 1;
while(p){
if(p & 1) res = (res * a) % mod;
a = (a * a) % mod;
p >>= 1;
}
return res;
}
inline void prime(){
for(int i = 2; i <= 300; i++){
for(int j = 1; j <= pcnt; j++)
if(i % p[j] == 0) pri[i] |= pri[ p[j] ];
if(pri[i] == 0) {p[++pcnt] = i; pri[i] = (1LL << (pcnt - 1LL));}
}
}
inline void get_inv(){
for(int i = 1; i <= pcnt; i++)
inv[i] = ksm(p[i], mod - 2);
}
// Segment Tree Start
struct node{
long long val,pri;
void operator+= (const node &b){
this -> val = (this -> val * b.val) % mod;
pri |= b.pri;
}
}tree[N << 2], lazy[N << 2];
inline void pushup(int now){
tree[now].val = (tree[now << 1].val * tree[now << 1 | 1].val) % mod;
tree[now].pri = tree[now << 1].pri | tree[now << 1 | 1].pri;
}
inline void pushdown(int now, int lson, int rson){
if(lazy[now].pri != 0){
tree[now << 1].val = (1LL * tree[now << 1].val * ksm(lazy[now].val, lson)) % mod;
tree[now << 1 | 1].val = (1LL * tree[now << 1 | 1].val * ksm(lazy[now].val, rson)) % mod;
tree[now << 1].pri |= lazy[now].pri;
tree[now << 1 | 1].pri |= lazy[now].pri;
lazy[now << 1].val = (1LL * lazy[now << 1].val * lazy[now].val) % mod;
lazy[now << 1 | 1].val = (1LL * lazy[now << 1 | 1].val * lazy[now].val) % mod;
lazy[now << 1].pri |= lazy[now].pri;
lazy[now << 1 | 1].pri |= lazy[now].pri;
lazy[now].val = 1LL; lazy[now].pri = 0;
}
}
inline void query_mut(int now, int left, int rig, int from, int to, int val){
if(from <= left && rig <= to){
tree[now].val = (1LL * tree[now].val * ksm(val, (rig - left + 1LL))) % mod;
tree[now].pri |= pri[val];
lazy[now].val = (lazy[now].val * val) % mod;
lazy[now].pri |= pri[val];
return ;
}
int mid = (left + rig) >> 1;
pushdown(now, mid - left + 1LL, rig - mid);
if(from <= mid) query_mut(now << 1, left, mid, from, to, val);
if(to > mid) query_mut(now << 1 | 1, mid + 1, rig, from, to, val);
pushup(now);
}
inline node query_sum(int now, int left, int rig, int from, int to){
if(from <= left && rig <= to) return tree[now];
int mid = (left + rig) >> 1;
node res = (node){1, 0};
pushdown(now, mid - left + 1LL, rig - mid);
if(from <= mid) res += query_sum(now << 1, left, mid, from, to);
if(to > mid) res += query_sum(now << 1 | 1, mid + 1, rig, from, to);
pushup(now);
return res;
}
inline void build_tree(int now, int left, int rig){
lazy[now].val = 1; lazy[now].pri = 0;
if(left == rig){
scanf("%lld", &tree[now].val);
tree[now].pri = pri[ tree[now].val ];
return ;
}
int mid = (left + rig) >> 1;
build_tree(now << 1, left, mid);
build_tree(now << 1 | 1, mid + 1, rig);
pushup(now);
}
// Segment Tree End
int main(){
prime();
get_inv();
scanf("%d%d", &n, &q);
build_tree(1, 1, n);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
while(q--){
scanf("%s", op);
if(op[0] == 'M'){
scanf("%d%d%d", &x, &y, &z);
query_mut(1, 1, n, x, y, z);
}
else {
scanf("%d%d", &x, &y);
node tmp = query_sum(1, 1, n, x, y);
for(int i = 1; i <= pcnt; i++){
if((tmp.pri >> (i - 1LL)) & 1LL)
tmp.val = (((tmp.val * (p[i] - 1) ) % mod) * inv[i]) % mod;
}
printf("%lld\n", tmp.val);
}
}
}]]>