太水，简述一下题意
T1
让你计算一个形如Σai * bi^ki
快速幂即可
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstdlib> 5 #include <cstring> 6 #include <ctime> 7 #define min(a, b) ((a) < (b) ? (a) : (b)) 8 #define max(a, b) ((a) > (b) ? (a) : (b)) 9 10 inline void swap(long long &x, long long &y) 11 { 12 long long tmp = x;x = y;y = tmp; 13 } 14 15 inline void read(long long &x) 16 { 17 x = 0;char ch = getchar(), c = ch; 18 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 19 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 20 if(c == '-')x = -x; 21 } 22 23 const long long INF = 0x3f3f3f3f; 24 const long long MAXN = 1000000; 25 26 long long n,k,MOD,ans,x; 27 28 long long pow(long long a, long long b) 29 { 30 long long r = 1, base = a; 31 for(;b;b >>= 1) 32 { 33 if(b & 1) r *= base, r %= MOD; 34 base *= base, base %= MOD; 35 } 36 return r; 37 } 38 39 long long tmp[100000], tot, a[MAXN], b[MAXN]; 40 41 int main() 42 { 43 freopen("digits.in", "r", stdin); 44 freopen("digits.out", "w", stdout); 45 read(n), read(k); 46 MOD = 1; 47 for(register long long i = 1;i <= k;++ i) MOD *= 10; 48 for(register long long i = 1;i <= n;++ i) 49 read(a[i]), read(b[i]); 50 read(x); 51 for(register long long i = 1;i <= n;++ i) 52 { 53 ans += (pow(x, b[i]) * a[i])%MOD, ans %= MOD; 54 } 55 for(register long long i = 1;i <= k;++ i) 56 { 57 tmp[++tot] = ans % 10; 58 ans /= 10; 59 } 60 for(register long long i = k;i >= 1;-- i) 61 printf("%d ", tmp[i]); 62 return 0; 63 }
T2
让你求a1 * x1 - x2 * x2 + x3 * x3 - a4 * x4 + a5 * x5 - a6 * x6 = 0在(0,k]内的解的个数，ai <= 10^5, k <= 600
meet in the middle可过，map做
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstdlib> 5 #include <cstring> 6 #include <ctime> 7 #include <map> 8 #define min(a, b) ((a) < (b) ? (a) : (b)) 9 #define max(a, b) ((a) > (b) ? (a) : (b)) 10 11 inline void swap(long long &x, long long &y) 12 { 13 long long tmp = x;x = y;y = tmp; 14 } 15 16 inline void read(long long &x) 17 { 18 x = 0;char ch = getchar(), c = ch; 19 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 20 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 21 if(c == '-')x = -x; 22 } 23 24 const long long INF = 0x3f3f3f3f; 25 26 std::map<long long, long long> mp; 27 long long num1[4], num2[4], k, ans; 28 29 int main() 30 { 31 freopen("equation.in", "r", stdin); 32 freopen("equation.out", "w", stdout); 33 read(k); 34 for(register long long i = 1;i <= 6;++ i) 35 { 36 if(i & 1)read(num1[(i - 1)/2 + 1]); 37 else read(num2[(i - 1)/2 + 1]); 38 } 39 for(register long long i = 1;i <= k;++ i) 40 for(register long long j = 1;j <= k;++ j) 41 for(register long long q = 1;q <= k;++ q) 42 ++ mp[i * num1[1] + j * num1[2] + q * num1[3]]; 43 for(register long long i = 1;i <= k;++ i) 44 for(register long long j = 1;j <= k;++ j) 45 for(register long long q = 1;q <= k;++ q) 46 if(mp.count(i * num2[1] + j * num2[2] + q * num2[3])) 47 ans += mp[i * num2[1] + j * num2[2] + q * num2[3]]; 48 printf("%lld ", ans); 49 return 0; 50 }
T3
让你求图上1到n的路径上的最大权值最小是多少
最小生成树 + dfs骚操作
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 #define min(a, b) ((a) < (b) ? (a) : (b)) 7 #define max(a, b) ((a) > (b) ? (a) : (b)) 8 9 inline void swap(long long &x, long long &y) 10 { 11 long long tmp = x;x = y;y = tmp; 12 } 13 14 inline void read(long long &x) 15 { 16 x = 0;char ch = getchar(), c = ch; 17 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 18 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 19 if(c == '-')x = -x; 20 } 21 22 const long long INF = 0x3f3f3f3f; 23 const long long MAXN = 1000000 + 10; 24 25 long long u[MAXN], v[MAXN], w[MAXN], cnt[MAXN]; 26 27 struct Edge 28 { 29 long long u,v,w,next; 30 Edge(long long _u, long long _v, long long _next, long long _w){w = _w;u = _u, v = _v,next = _next;} 31 Edge(){} 32 }edge[MAXN << 1]; 33 long long head[MAXN], cntt; 34 35 inline void insert(long long a, long long b, long long c) 36 { 37 edge[++cntt] = Edge(a,b,head[a],c); 38 head[a] = cntt; 39 } 40 41 long long n,m,k,fa[MAXN]; 42 43 long long find(long long x) 44 { 45 return fa[x] == x ? x : fa[x] = find(fa[x]); 46 } 47 48 long long cmp(long long a, long long b) 49 { 50 return w[a] < w[b]; 51 } 52 53 long long ans, b[MAXN]; 54 55 long long dfs(long long u) 56 { 57 b[u] = 1; 58 for(register long long pos = head[u];pos;pos = edge[pos].next) 59 { 60 long long v = edge[pos].v; 61 if(v == n) 62 { 63 ans = max(ans, edge[pos].w); 64 return 1; 65 } 66 if(b[v]) continue; 67 if(dfs(v)) 68 { 69 ans = max(ans, edge[pos].w); 70 return 1; 71 } 72 } 73 return 0; 74 } 75 76 int main() 77 { 78 freopen("graph.in", "r", stdin); 79 freopen("graph.out", "w", stdout); 80 read(n), read(m), read(k); 81 for(register long long i = 1;i <= m;++ i) 82 read(u[i]), read(v[i]), read(w[i]), cnt[i] = i; 83 std::sort(cnt + 1, cnt + 1 + m, cmp); 84 for(register long long i = 1;i <= n;++ i) fa[i] = i; 85 for(register long long p = 1;p <= m;++ p) 86 { 87 long long i = cnt[p]; 88 long long f1 = find(u[i]), f2 = find(v[i]); 89 if(f1 == f2)continue; 90 fa[f1] = f2; 91 insert(u[i], v[i], w[i]); 92 insert(v[i], u[i], w[i]); 93 } 94 dfs(1); 95 if(ans == 0)ans = -1; 96 printf("%lld", ans); 97 return 0; 98 }