ARC061
C - たくさんの数式 / Many Formulas
这个其实\(10^5\)也能做。。
就是\(dp[i]\)表示到第i位的方案数，\(sum[i]\)表示延伸到第i位之前的所有方案的数字的和，\(pre[i]\)记录到第i位延伸已经结束了的数字的答案
转移是\(dp[i] = dp[i - 1] * 2\)
\(sum[i] = sum[i - 1] * 10 + dp[i - 1] * (s[i] - '0')\)
\(pre[i] = pre[i - 1] * 2 + sum[i ]\)
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
char s[15];
int64 dp[15],sum[15],ans,pre[15];
int N;
void Solve() {
scanf("%s",s + 1);
dp[0] = 1;
N = strlen(s + 1);
for(int i = 1 ; i <= N ; ++i) {
sum[i] = sum[i - 1] * 10 + (s[i] - '0') * dp[i - 1];
dp[i] = dp[i - 1] * 2;
if(i != N) pre[i] = pre[i - 1] * 2 + sum[i];
else pre[N] = pre[i - 1] + sum[i];
}
out(pre[N]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} D - すぬけ君の塗り絵 / Snuke's Coloring
认为一个\(3\times 3\)是中心格子八个方位加上自己，统计每个\(3\times 3\)在中心格子统计
发现一共有\((H-2)(W - 2)\)种，对于不为0的中心格子只可能在黑格子附近八个方位加黑格子自己，暴力统计即可
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int H,W,N;
int a[MAXN],b[MAXN];
int64 ans[MAXN];
map zz;
map cnt;
int dx[9] = {-1,1,0,0,0,1,1,-1,-1};
int dy[9] = {0,0,-1,1,0,1,-1,1,-1};
void Solve() {
read(H);read(W);read(N);
for(int i = 1 ; i <= N ; ++i) {
read(a[i]);read(b[i]);
zz[mp(a[i],b[i])] = 1;
}
for(int i = 1 ; i <= N ; ++i) {
for(int k = 0 ; k < 9 ; ++k) {
int tx = a[i] + dx[k];
int ty = b[i] + dy[k];
if(tx > 1 && tx < H && ty > 1 && ty < W) {
int c = 0;
for(int h = 0 ; h < 9 ; ++h) {
if(zz[mp(tx + dx[h],ty + dy[h])]) ++c;
}
cnt[mp(tx,ty)] = c;
}
}
}
ans[0] = 1LL * (H - 2) * (W - 2);
for(auto t : cnt) {
ans[0]--;
ans[t.se]++;
}
for(int i = 0 ; i <= 9 ; ++i) {
out(ans[i]);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} E - すぬけ君の地下鉄旅行 / Snuke's Subway Trip
把边建成点，一个点上连着的边同种颜色用长度为0的边连到一起，从每种颜色中选一个代表边，新建一个点往上面连去是0回来是1的边，表示花费1的代价在这个点转移到别的颜色的边
然后跑dij就行了
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 400005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct node {
int to,next,val;
}E[MAXN * 10];
int head[MAXN],sumE;
int N,M,Ncnt;
int c[MAXN],dis[MAXN];
vector to[MAXN];
bool vis[MAXN];
priority_queue Q;
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].val = c;
head[u] = sumE;
}
void Solve() {
read(N);read(M);
int q,p;
Ncnt = M;
for(int i = 1 ; i <= M ; ++i) {
read(q);read(p);read(c[i]);
to[q].pb(i);to[p].pb(i);
}
for(int i = 1 ; i <= N ; ++i) {
sort(to[i].begin(),to[i].end(),[](int a,int b) {return c[a] < c[b];});
int nw = ++Ncnt;
for(int j = 0 ; j < to[i].size() ; ++j) {
int p = j;
while(p < to[i].size() - 1 && c[to[i][p + 1]] == c[to[i][j]]) ++p;
for(int h = j + 1 ; h <= p ; ++h) {
add(to[i][j],to[i][h],0);
add(to[i][h],to[i][j],0);
}
add(to[i][j],nw,0);
add(nw,to[i][j],1);
j = p;
}
}
for(int i = 1 ; i <= Ncnt ; ++i) dis[i] = 1e9;
for(auto t : to[1]) {dis[t] = 1;Q.push(mp(-1,t));}
while(!Q.empty()) {
pii now = Q.top();Q.pop();
if(vis[now.se]) continue;
int u = now.se;vis[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(dis[v] > dis[u] + E[i].val) {
dis[v] = dis[u] + E[i].val;
Q.push(mp(-dis[v],v));
}
}
}
int ans = 1e9;
for(auto t : to[N]) ans = min(ans,dis[t]);
if(ans >= 1e9) {
puts("-1");return;
}
else {out(ans);enter;}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} F - 3人でカードゲーム / Card Game for Three
不同的玩家序列代表了不同的分牌方式，只不过同种玩家序列可能有很多中分牌方式，这种不同是输的两个玩家手中剩余的牌不同的可能性带来的
于是我们发现，A必胜的玩家序列是一个\(L >= N + 1\)的序列，其中两头都是a，中间有\(N - 1\)个a，和\((L - 1 - N)\)个\(b\)或\(c\)，他们各自不超过自己的上限\(M\)和\(K\)
于是我们发现如果有\(n\)个\(b\)或\(c\)
他们合法的排列是一段区间\([l,r]\)里的\(\sum_{i = l}^{r}\binom{i}{n}\)
而i变大1时，\([l,r]\)两个端点的变化不会超过1
发现我们可以利用\(n\)的\([l,r]\)的组合数的和，快速算出\(n + 1\)的\([l + 1,r+ 1]\)的组合数的和，然后我们只需要修改左右端点不合法的情况使区间合法即可
复杂度\(O(M+ K)\)
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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,M,K,fac[MAXN],invfac[MAXN],pw[MAXN];
int s[MAXN];
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 C(int n,int m) {
if(n < m) return 0;
else return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
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;
}
void Solve() {
read(N);read(M);read(K);
fac[0] = 1;
for(int i = 1 ; i <= 1000000 ; ++i) fac[i] = mul(fac[i - 1],i);
invfac[1000000] = fpow(fac[1000000],MOD - 2);
for(int i = 999999 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
pw[0] = 1;
for(int i = 1 ; i <= 1000000 ; ++i) pw[i] = mul(pw[i - 1],3);
int l = 0,r = 0;s[0] = 1;
for(int i = 1 ; i <= M + K ; ++i) {
int tmp = inc(mul(s[i - 1],2),inc(C(i - 1,r + 1),MOD - C(i - 1,l)));
++r;++l;
while(r + 1 <= i && r + 1 <= M) {update(tmp,C(i,r + 1));++r;}
while(r > M) {update(tmp,MOD - C(i,r));--r;}
while(i - l > K) {update(tmp,MOD - C(i,l));++l;}
while(i - (l - 1) <= K && (l - 1) >= 0) {update(tmp,C(i,l - 1));--l;}
s[i] = tmp;
}
int ans = 0;
for(int i = 0 ; i <= M + K ; ++i) {
int t = mul(C(N - 1 + i,N - 1),s[i]);
t = mul(t,pw[M + K - i]);
update(ans,t);
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}