ARC064
C - Boxes and Candies
先把每个盒子都消到x
然后从前往后推，要求第二个的上界是x-前一个
因为我们要求靠后的那个尽量小，会对后面的修改影响尽量小
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int64 a[MAXN],x,ans;
void Solve() {
read(N);read(x);
for(int i = 1 ; i <= N ; ++i) {
read(a[i]);
if(a[i] > x) {
ans += a[i] - x;
a[i] = x;
}
}
int64 t = 0;
for(int i = 2 ; i <= N ; ++i) {
int64 m = x - a[i - 1];
if(a[i] > m) {t += a[i] - m;a[i] = m;}
}
out(ans + t);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
D - An Ordinary Game
最后的串长度的奇偶性是一定的
例如两端是a和c，最后的串一定是acac，或者acacacac
两端是a的话最后的一定是aba，或者ababa，b可以换成任意字母
我们看看两个串的长度的差值是奇数还是偶数
然后根据奇偶性判断胜负即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
char s[MAXN];
int L;
void Solve() {
scanf("%s",s + 1);
L = strlen(s + 1);
if(s[1] == s[L]) {
--L;
if(L & 1) puts("First");
else puts("Second");
}
else {
if(L & 1) puts("First");
else puts("Second");
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
E - Cosmic Rays
就是跑一遍dij就行，两个圆之间的距离要么是0要么是圆心距减两个半径
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 1005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct Point {
db x,y;
Point(db _x = 0.0,db _y = 0.0) {x = _x;y = _y;}
friend Point operator + (const Point &a,const Point &b) {
return Point(a.x + b.x,a.y + b.y);
}
friend Point operator - (const Point &a,const Point &b) {
return Point(a.x - b.x,a.y - b.y);
}
db norm() {
return sqrt(x * x + y * y);
}
}P[MAXN],S,T;
db R[MAXN],dis[MAXN];
int N;
bool vis[MAXN];
void Solve() {
scanf("%lf%lf%lf%lf",&S.x,&S.y,&T.x,&T.y);
read(N);
for(int i = 1 ; i <= N ; ++i) {
scanf("%lf%lf%lf",&P[i].x,&P[i].y,&R[i]);
dis[i] = max(0.0,(S - P[i]).norm() - R[i]);
}
for(int i = 1 ; i <= N ; ++i) {
int u = -1;
for(int j = 1 ; j <= N ; ++j) {
if(!vis[j]) {
if(u == -1) u = j;
else if(dis[j] < dis[u]) u = j;
}
}
vis[u] = 1;
for(int j = 1 ; j <= N ; ++j) {
if(!vis[j]) {
dis[j] = min(dis[j],dis[u] + max(0.0,(P[u] - P[j]).norm() - R[u] - R[j]));
}
}
}
db ans = (S - T).norm();
for(int i = 1 ; i <= N ; ++i) {
ans = min(ans,dis[i] + max(0.0,(P[i] - T).norm() - R[i]));
}
printf("%.10lf
",ans);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
F - Rotated Palindromes
记(f(i))为长度为i个回文串个数，不含循环节
(f(i) = K^{lceil frac{i}{2} ceil} - sum_{d|i}f(d))
由于i只有(O(sqrt{N}))个且转移关系并不多，所以(f(i))都可以很快求出来
然后我们答案就是
(sum_{d|N}dcdot f(d)[d \%2 == 1] + sum_{d|N}frac{d}{2} cdot f(d)[d \% 2 == 0])
因为d如果是2的倍数，转一半的时候会构成一个新的d长度回文串，重复统计了
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 1005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,K;
map<int,int> zz;
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &x,int y) {
x = inc(x,y);
}
int fpow(int x,int c) {
int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
int f(int x) {
if(x == 1) return K;
if(zz.count(x)) return zz[x];
int res = fpow(K,x / 2);
if(x & 1) res = mul(res,K);
for(int i = 1 ; i <= x / i ; ++i) {
if(x % i == 0) {
update(res,MOD - f(i));
int j = x / i;
if(j != i && j != x) update(res,MOD - f(j));
}
}
zz[x] = res;
return res;
}
void Solve() {
read(N);read(K);
int ans = 0;
for(int i = 1 ; i <= N / i; ++i) {
if(N % i == 0) {
int t = mul(f(i),i);
if(i % 2 == 0) t = mul(t,(MOD + 1) / 2);
update(ans,t);
int j = N / i;
t = mul(f(j),j);
if(j % 2 == 0) t = mul(t,(MOD + 1) / 2);
if(j != i) update(ans,t);
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}