标签：signed back mod 情况 spl pair for bad 三角形
A. 2:40:11(-7) solved by zcz
通过旋转使得抓人的在左下角，逃得在右上角
结论是逃得一定在右上 或者 右下 左上被抓住，找到规律枚举一下即可
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define LL long long long long n,m; void change(long long &x,long long &y) { long long t=x; x=m+1-y; y=t; } const int now_size = 3; const int mod = 1025436931; class Matix { public: LL mdata[3][3]; Matix(int value = 0) { memset(mdata, 0, sizeof(mdata)); if (value == 1) { for (int i = 0; i < now_size; i++) mdata[i][i] = 1; } } Matix(const Matix &s1) { for (int i = 0; i < now_size; i++) { for (int j = 0; j < now_size; j++) { mdata[i][j] = s1.mdata[i][j]; } } } Matix operator * (const Matix &s1) { Matix ans; for (int i=0; i<now_size; i++) { for (int j=0; j<now_size; j++) { for (int k=0; k<now_size; k++) { ans.mdata[i][j] = (ans.mdata[i][j] + mdata[i][k]* s1.mdata[k][j])%mod; } } } return ans; } }; LL fbnq(LL k) { Matix p1; p1.mdata[0][0] = 1; p1.mdata[0][1] = 1; p1.mdata[0][2] = 1; p1.mdata[1][1] = 1; p1.mdata[1][2] = 1; p1.mdata[2][1] = 1; Matix ans(1); k--; while (k) { if (k&1) ans = ans *p1; p1 = p1*p1; k>>=1; } LL fans = (ans.mdata[0][0] + ans.mdata[0][1]) % mod; return fans; } int main() { int T; cin>>T; while(T--) { long long x1,x2,y1,y2; cin>>n>>m>>x1>>y1>>x2>>y2; if(n==1&&m==1) { cout<<"countless"<<endl; continue; } if(x1==x2&&y1==y2) { cout<<0<<endl; continue; } while(!(x1<x2&&y1<=y2)) { change(x1,y1); change(x2,y2); swap(n,m); } if(m==1) { if((x2-x1)%2==0) { cout<<fbnq(n-x1-1)<<endl; } else { cout<<fbnq(n-x1)<<endl; } continue; } if((x2-x1+y2-y1)%2==1) { cout<<"countless"<<endl; continue; } long long ans=n-x1+m-y1; if(x2-x1>y2-y1) { ans=max(ans,n-x1+y1-1); } if(x2-x1<y2-y1) { ans=max(ans,m-y1+x1-1); } cout<<fbnq(ans-1)<<endl; } return 0; }
B.1:28:35(-1) solved by zcz
并查集缩个点 dfs计算贡献
#include<iostream> #include<cstring> #include<cstdio> #include<vector> using namespace std; vector<int> e[100005]; int M[100005]; int pre[100005]; int cnt[100005]; int Find(int x) { if(x!=pre[x]) { pre[x]=Find(pre[x]); } return pre[x]; } int main() { int T; cin>>T; while(T--) { int n,m,k; cin>>n>>m>>k; for(int i=1;i<=n;i++) { e[i].clear(); M[i]=0; pre[i]=i; cnt[i]=1; } while(m--) { int a,b; scanf("%d %d",&a,&b); e[a].push_back(b); e[b].push_back(a); } while(k--) { int a; scanf("%d",&a); M[a]=1; } for(int i=1;i<=n;i++) { for(int j=0;j<e[i].size();j++) { if(!M[i]&&!M[e[i][j]]) { int x=Find(i); int y=Find(e[i][j]); if(x!=y) { pre[x]=y; cnt[y]+=cnt[x]; } } } } int p=Find(1); double ans=0; for(int i=1;i<=n;i++) { if(M[i]) { int c2=0; long long c1=0; int l=e[i].size(); for(int j=0;j<l;j++) { if(M[e[i][j]]) continue; if(Find(e[i][j])==p) { c2++; } else { c1+=cnt[Find(e[i][j])]; } } if(c2>0&&c2<l) { ans=max(ans,1.0*c1/l); } } } printf("%.8lf\n",cnt[p]+ans); } return 0; }
C.00:44:49(-1) solved by hl
模型都不改的多重背包
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; PII p[maxn]; LL dp[maxn]; bool cmp(PII a,PII b){ return a.se < b.se; } int main(){ while(~Sca2(N,M)){ for(int i = 1; i <= N ; i ++){ Sca2(p[i].fi,p[i].se); } sort(p + 1,p + 1 + N,cmp); int q = M + 1e4 + 1; for(int i = 0 ; i <= q; i ++) dp[i] = 1e18; dp[0] = 0; for(int i = 0; i <= q; i ++){ for(int j = 1; j <= N ; j ++){ if(i >= p[j].se) dp[i] = min(dp[i],dp[i - p[j].se] + p[j].fi); else dp[i] = min(dp[i],dp[0] + p[j].fi); } } LL ans = dp[M]; while(dp[M] == dp[M + 1]) M++; printf("%lld %d\n",ans,M); } return 0; }
D.1:07:35 solved by hl
点分治,维护子树余数为0,1,2的长度和以及点的数量，直接计算贡献即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; LL dis; }edge[maxn << 1]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v,LL w){ edge[tot].to = v; edge[tot].next = head[u]; edge[tot].dis = w; head[u] = tot++; } int root,max_part,SUM; bool vis[maxn]; int size[maxn]; void dfs_root(int u,int la){ size[u] = 1;int heavy = 0; for(int i = head[u]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la || vis[v]) continue; dfs_root(v,u); heavy = max(heavy,size[v]); size[u] += size[v]; } if(max_part > max(heavy,SUM - heavy)){ max_part = max(heavy,SUM - heavy); root = u; } } LL A,B,C; LL distmp[5],dis[5],num[5],numtmp[5]; void dfs_dis(int t,int la,LL d){ numtmp[d % 3]++; distmp[d % 3] += d; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(vis[v] || v == la) continue; dfs_dis(v,t,d + edge[i].dis); } } void work(int t){ num[0]++; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(vis[v]) continue; dfs_dis(v,t,edge[i].dis); for(int k = 0 ; k < 3; k ++){ for(int j = 0 ; j < 3; j ++){ if((k + j) % 3 == 0){ A = (A + numtmp[k] * dis[j]) % mod; A = (A + num[k] * distmp[j]) % mod; }else if((k + j) % 3 == 1){ B = (B + numtmp[k] * dis[j]) % mod; B = (B + num[k] * distmp[j]) % mod; }else{ C = (C + numtmp[k] * dis[j]) % mod; C = (C + num[k] * distmp[j]) % mod; } } } for(int k = 0 ; k < 3; k ++){ dis[k] += distmp[k]; distmp[k] = 0; dis[k] %= mod; num[k] += numtmp[k]; numtmp[k] = 0; num[k] %= mod; } } for(int i = 0 ; i < 3; i ++) dis[i] = num[i] = 0; } void divide(int t){ max_part = INF,root = t; dfs_root(t,-1); vis[root] = 1; work(root); for(int i = head[root]; ~i; i = edge[i].next){ int v = edge[i].to; if(vis[v]) continue; SUM = size[v]; divide(v); } } int main(){ while(~Sca(N)){ init(); A = B = C = 0; for(int i = 0 ; i <= N ; i ++) size[i] = vis[i] = 0; for(int i = 1; i < N ; i ++){ int u = read() + 1,v = read() + 1; LL w = read(); add(u,v,w); add(v,u,w); } SUM = N; divide(1); printf("%lld %lld %lld\n",(A * 2) % mod,(B * 2) % mod,(C * 2) % mod); } return 0; }
F.12:25:23 solved by hl
二分枚举最大最小值，特判最大值小于等于最小值的情况即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 5e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL a[maxn]; bool check(LL x){ LL ans = 0; for(int i = 1; i <= N ; i ++) if(a[i] > x) ans += a[i] - x; return ans <= K; } LL solve(){ LL l = 0,r = 1e9; LL ans = 1e9; while(l <= r){ LL m = l + r >> 1; if(check(m)){ r = m - 1; ans = m; }else{ l = m + 1; } } return ans; } bool check2(LL x){ LL ans = 0; for(int i = 1; i <= N ; i ++) if(a[i] < x) ans += x - a[i]; return ans <= K; } LL solve2(){ LL l = 0,r = 1e9; LL ans = 0; while(l <= r){ LL m = l + r >> 1; if(check2(m)){ l = m + 1; ans = m; }else{ r = m - 1; } } return ans; } int main(){ while(~Sca2(N,K)){ LL sum = 0; for(int i = 1; i <= N ; i ++) sum += a[i] = read(); LL r = solve(),l = solve2(); if(r <= l){ if(sum % N) puts("1"); else puts("0"); }else{ Prl(r - l); } } return 0; }
G.4:08:33(-3) solved by zcz
利用三角形等效变换推来推去推出公式，然后求一个极限，数字过大的时候直接输出极限
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; char s[1000005]; int main() { int T; cin>>T; while(T--) { long long n; double a; scanf("%s",s); cin>>a; double ans=2; double x=ans; if(strlen(s)<7) { sscanf(s,"%lld",&n); n--; while(n--) { ans=ans/(1+3*ans)+5.0/3; } } else { ans=(sqrt(5.0)+1)/2+1.0/3; } printf("%.10lf\n",(ans-1.0/3)*a); } return 0; }
H.1:49:00 solved by hl
毫无坑点模拟题
硬说有坑点的话，五个头不算炸，算四带一
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; string t; struct node{ int a,b,c,d; node(int a = 0,int b = 0,int c = 0,int d = 0):a(a),b(b),c(c),d(d){} }; struct P{ string name; node val; }p[maxn]; int a[10]; map<char,int>Q; void init(){ for(int i = 2; i <= 9; i ++) Q[i + ‘0‘] = i; Q[‘A‘] = 1; Q[‘0‘] = 10; Q[‘J‘] = 11; Q[‘Q‘] = 12; Q[‘K‘] = 13; } node solve(string t){ int cnt = 0; for(int i = t.size() - 1; i >= 0 ; i --){ a[++cnt] = Q[t[i]]; if(t[i] == ‘0‘) i--; } sort(a + 1,a + 1 + cnt); if(a[1] == 1 && a[2] == 10 && a[3] == 11 && a[4] == 12 && a[5] == 13) return node(9,1); if(a[1] + 1 == a[2] && a[2] + 1 == a[3] && a[3] + 1 == a[4] && a[4] + 1 == a[5]) return node(8,a[5]); //if(a[1] == a[5]) return node(7,a[1]); if(a[1] == a[4]) return node(6,a[1],a[5]); if(a[2] == a[5]) return node(6,a[2],a[1]); if(a[1] == a[3] && a[4] == a[5]) return node(5,a[1],a[4]); if(a[3] == a[5] && a[1] == a[2]) return node(5,a[3],a[1]); if(a[1] == a[3]) return node(4,a[1],a[4] + a[5]); if(a[2] == a[4]) return node(4,a[2],a[1] + a[5]); if(a[3] == a[5]) return node(4,a[3],a[1] + a[2]); if(a[2] == a[3] && a[4] == a[5]) return node(3,a[5],a[3],a[1]); if(a[1] == a[2] && a[4] == a[5]) return node(3,a[5],a[2],a[3]); if(a[1] == a[2] && a[3] == a[4]) return node(3,a[3],a[1],a[5]); if(a[1] == a[2]) return node(2,a[2],a[3] + a[4] + a[5]); if(a[2] == a[3]) return node(2,a[2],a[1] + a[4] + a[5]); if(a[3] == a[4]) return node(2,a[3],a[1] + a[2] + a[5]); if(a[4] == a[5]) return node(2,a[4],a[1] + a[2] + a[3]); return node(1,a[1] + a[2] + a[3] + a[4] + a[5]); } bool cmp(P a,P b){ node x = a.val,y = b.val; if(x.a != y.a) return x.a > y.a; if(x.b != y.b) return x.b > y.b; if(x.c != y.c) return x.c > y.c; if(x.d != y.d) return x.d > y.d; return a.name < b.name; } int main(){ init(); while(~Sca(N)){ for(int i = 1; i <= N ; i ++){ cin >> p[i].name; string tmp; cin >> tmp; p[i].val = solve(tmp); // cout << p[i].val.a << endl; } sort(p + 1,p + 1 + N,cmp); for(int i = 1; i <= N ; i ++){ cout << p[i].name << endl; } } return 0; }
J.2:40:35(-1) solved by gbs
队友过的
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; const int maxn = 1e6+15; const int mod = 1e9 + 7; //快速幂 long long pows[maxn+5]; long long unko[maxn+5]; long long w[maxn+5]; long long C(int n,int m) { return pows[n]*unko[m]%mod*unko[n-m]%mod; } int quick_mi(LL a, LL k) { LL ans = 1; while (k) { if (k & 1) { ans *= a; ans %= mod; } a *= a; a %= mod; k >>= 1; } return ans; } LL inva(int a) { return quick_mi(a,mod-2); } LL rm[maxn+15]; LL cn[maxn+15]; int main() { pows[0]=1;pows[1]=1; unko[0]=1;unko[1]=1; w[1]=1; for(int i = 2;i<maxn;i++) { w[i]=mod-(long long)(mod/i)*w[mod%i]%mod; pows[i]=pows[i-1]*i%mod; unko[i]=unko[i-1]*w[i]%mod; } rm[0] =1; cn[0] = 1; rm[1] =1; rm[2] =2; cn[1] = 1; for (int i=3; i<maxn; i++) { rm[i] = (rm[i-1] *i)%mod; } for (int i=2; i<maxn; i++) { cn[i] = rm[i-1]; } int T; int N,X; cin >>T; while(T--) { scanf("%d%d",&N,&X); LL b1 =rm[N]; LL a1 =rm[N]; LL fz = 1; LL fm = 1; for (int i=1; i<=N; i++) { fz = (fz * (N-i+1))%mod; fm = (fm * i)%mod; if (i>X) { a1 = a1 - (((C(N,i) * cn[i])%mod) * rm[N-i] )%mod ; a1 = (a1%mod+mod)%mod; } } printf("%lld\n",(a1 *inva(b1))%mod); } #ifdef VSCode system("pause"); #endif return 0; }
K.1:26:14 (-2) solved by gbs
队友过的
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; const int mod = 1e9+7; int now_size; int gaosi[100][100]; int n; LL k; int an[200]; LL suman[200]; LL fin_ans[100]; class Matix { public: LL mdata[75][75]; Matix(int value = 0) { memset(mdata, 0, sizeof(mdata)); if (value == 1) { for (int i = 0; i < now_size; i++) mdata[i][i] = 1; } } Matix(const Matix &s1) { for (int i = 0; i < now_size; i++) { for (int j = 0; j < now_size; j++) { mdata[i][j] = s1.mdata[i][j]; } } } Matix operator * (const Matix &s1) { Matix ans; for (int i=0; i<now_size; i++) { for (int j=0; j<now_size; j++) { for (int k=0; k<now_size; k++) { ans.mdata[i][j] = (ans.mdata[i][j] + mdata[i][k]* s1.mdata[k][j])%mod; } } } return ans; } void out () { for (int i=0; i<now_size; i++) { for (int j=0; j<now_size ;j++) { cout<<mdata[i][j]<<‘\t‘; } cout<<endl; } cout<<endl; } }; int quick_mi(LL a, LL k) { LL ans = 1; while (k) { if (k & 1) { ans *= a; ans %= mod; } a *= a; a %= mod; k >>= 1; } return ans; } inline int inva(int a) { return quick_mi(a,mod-2); } inline int domod(int a) { return (a%mod+mod)%mod; } inline LL domod(LL a) { return (a%mod+mod)%mod; } void gaosixiaoyuan(int hang,int lie) { for (int i =0; i<n; i++) { //cout <<i <<"^"<<endl; for (int j = i; j< hang; j++) { if (gaosi[j][i] != 0) { if (i != j) { for (int k1 = 0; k1<lie; k1++) swap(gaosi[j][k1],gaosi[i][k1]); } break; } } int qmod = inva(gaosi[i][i]); //cout<<qmod<<"*"<<endl; for (int k1 = 0; k1<lie; k1++) gaosi[i][k1] = domod(gaosi[i][k1] *1LL *qmod); for (int j= i+1; j<hang ;j++) { if (gaosi[j][i] != 0) { int p1 = gaosi[j][i]; for (int k1 = i; k1<lie; k1++) gaosi[j][k1] = domod(gaosi[j][k1] -1LL * p1 *gaosi[i][k1]); } } } for (int i= n-1; i>=0; i--) { LL nowans = gaosi[i][n]; for (int j = n-1; j>i; j--) { nowans = domod(nowans - fin_ans[j] * gaosi[i][j]); } fin_ans[i] = nowans; } /*for (int i=0; i<hang; i++) { for (int j=0; j<lie ;j++) { cout<<gaosi[i][j]<<‘\t‘; } cout<<endl; } cout<<endl; cout<<"%%"; for (int j=0; j<n ;j++) { cout<<fin_ans[j]<<‘\t‘; } cout<<endl;*/ } int main() { int t; //cout<<(2*inva(2))%mod<<endl; cin >> t; while (t--) { cin >> n >> k; for (int i = 0; i < 2 * n; i++) { scanf("%d", &an[i]); if (i==0) { suman[i] = an[i]; } else { suman[i] = domod(an[i] +suman[i-1]); } } if (k<=2*n) { k--; printf("%lld\n",suman[k]); continue; } for (int i = 0; i < n; i++) { for (int j = 0; j <= n; j++) { gaosi[i][j] = an[i + j]; } } for (int j = 0; j <= n; j++) gaosi[n][j] = 1; gaosixiaoyuan(n+1,n+1); now_size = n+2; Matix simp1; simp1.mdata[0][0] = 1; //simp1.mdata[0][1] = 1; for (int i=0; i<n; i++) { simp1.mdata[0][i+1] = fin_ans[n-1-i]; simp1.mdata[1][i+1] = fin_ans[n-1-i]; } for (int i =2 ; i<=n+1; i++) { simp1.mdata[i][i-1] = 1; } //simp1.out(); Matix ans1(1); k -= n; while(k) { if (k&1) { ans1 = ans1 *simp1; } k>>=1; simp1 = simp1 *simp1; //simp1.out(); } LL the_ans = suman[n-1]; //cout<<"&"<<the_ans<<endl; for (int i = 1; i<=n; i++) { the_ans = domod(the_ans + an[n-i] *ans1.mdata[0][i]); } printf("%lld\n",the_ans); } #ifdef VSCode system("pause"); #endif return 0; }
标签：signed back mod 情况 spl pair for bad 三角形
原文地址：https://www.cnblogs.com/Hugh-Locke/p/11519568.html
