这套题感觉出的不咋滴,第四题和第五题难度差了1000分!!!
前四题都还简单,第五题就31人做出……我算了……
懒得写题解了,做个记录吧(这就是偷懒的理由???)
A.New Year Garland
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #define rep(x, l, r) for(int x = l; x <= r; x++) 7 #define repd(x, r, l) for(int x = r; x >= l; x--) 8 #define clr(x, y) memset(x, y, sizeof(x)) 9 #define all(x) x.begin(), x.end() 10 #define pb push_back 11 #define mp make_pair 12 #define MAXN 13 #define fi first 14 #define se second 15 #define SZ(x) ((int)x.size()) 16 using namespace std; 17 typedef long long ll; 18 typedef vector<int> vi; 19 typedef pair<int, int> pii; 20 const int INF = 1 << 30; 21 const int p = 1000000009; 22 int lowbit(int x){ return x & (-x);} 23 int fast_power(int a, int b){ int x; for(x = 1; b; b >>= 1){ if(b & 1) x = 1ll * x * a % p; a = 1ll * a * a % p;} return x % p;} 24 25 int main(){ 26 int t; 27 scanf("%d", &t); 28 while(t--){ 29 int a, b, c; 30 scanf("%d%d%d", &a, &b, &c); 31 if(a < b) swap(a, b); 32 if(a < c) swap(a, c); 33 if(a - 1 > b + c) puts("No"); 34 else puts("Yes"); 35 } 36 return 0; 37 }
-A
B.Verse For Santa
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #define rep(x, l, r) for(int x = l; x <= r; x++) 7 #define repd(x, r, l) for(int x = r; x >= l; x--) 8 #define clr(x, y) memset(x, y, sizeof(x)) 9 #define all(x) x.begin(), x.end() 10 #define pb push_back 11 #define mp make_pair 12 #define MAXN 100005 13 #define fi first 14 #define se second 15 #define SZ(x) ((int)x.size()) 16 using namespace std; 17 typedef long long ll; 18 typedef vector<int> vi; 19 typedef pair<int, int> pii; 20 const int INF = 1 << 30; 21 const int p = 1000000009; 22 int lowbit(int x){ return x & (-x);} 23 int fast_power(int a, int b){ int x; for(x = 1; b; b >>= 1){ if(b & 1) x = 1ll * x * a % p; a = 1ll * a * a % p;} return x % p;} 24 25 int a[MAXN]; 26 27 int main(){ 28 int t; 29 scanf("%d", &t); 30 rep(times, 1, t){ 31 int n, s; 32 scanf("%d%d", &n, &s); 33 rep(i, 1, n) scanf("%d", &a[i]); 34 int sum = 0, maxx = 0, ans = 0; 35 rep(i, 1, n){ 36 sum += a[i]; 37 if(a[i] > maxx){ 38 maxx = a[i]; 39 ans = i; 40 } 41 if(sum > s) break; 42 } 43 if(sum <= s) puts("0"); 44 else printf("%d\n", ans); 45 } 46 return 0; 47 }
-B
C.Stack of Presents
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #define rep(x, l, r) for(int x = l; x <= r; x++) 7 #define repd(x, r, l) for(int x = r; x >= l; x--) 8 #define clr(x, y) memset(x, y, sizeof(x)) 9 #define all(x) x.begin(), x.end() 10 #define pb push_back 11 #define mp make_pair 12 #define MAXN 13 #define fi first 14 #define se second 15 #define SZ(x) ((int)x.size()) 16 using namespace std; 17 typedef long long ll; 18 typedef vector<int> vi; 19 typedef pair<int, int> pii; 20 const int INF = 1 << 30; 21 const int p = 1000000009; 22 int lowbit(int x){ return x & (-x);} 23 int fast_power(int a, int b){ int x; for(x = 1; b; b >>= 1){ if(b & 1) x = 1ll * x * a % p; a = 1ll * a * a % p;} return x % p;} 24 25 int main(){ 26 int t; 27 scanf("%d", &t); 28 rep(times, 1, t){ 29 int n; 30 scanf("%d", &n); 31 ll s1 = 0, s2 = 0; 32 rep(i, 1, n){ 33 ll x; 34 scanf("%lld", &x); 35 s1 += x; 36 s2 ^= x; 37 } 38 s2 <<= 1; 39 ll ans = 0; 40 for(int i = 0; s1 != s2; i++){ 41 if((s1 & 1) ^ (s2 & 1)){ 42 s1 += 1; 43 s2 ^= 2; 44 ans += 1ll << i; 45 } 46 s1 >>= 1; 47 s2 >>= 1; 48 } 49 printf("1\n%lld\n", ans); 50 } 51 return 0; 52 }
-C
D.Santa‘s Bot
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #define rep(x, l, r) for(int x = l; x <= r; x++) 7 #define repd(x, r, l) for(int x = r; x >= l; x--) 8 #define clr(x, y) memset(x, y, sizeof(x)) 9 #define all(x) x.begin(), x.end() 10 #define pb push_back 11 #define mp make_pair 12 #define MAXN 1000005 13 #define fi first 14 #define se second 15 #define SZ(x) ((int)x.size()) 16 using namespace std; 17 typedef long long ll; 18 typedef vector<int> vi; 19 typedef pair<int, int> pii; 20 const int INF = 1 << 30; 21 const int p = 998244353; 22 int lowbit(int x){ return x & (-x);} 23 int fast_power(int a, int b){ int x; for(x = 1; b; b >>= 1){ if(b & 1) x = 1ll * x * a % p; a = 1ll * a * a % p;} return x % p;} 24 25 vi a[MAXN]; 26 int inv[MAXN], id[MAXN], sum[MAXN]; 27 28 int main(){ 29 int n; 30 scanf("%d", &n); 31 inv[1] = 1; 32 rep(i, 2, 1000000) inv[i] = 1ll * (p - p / i) * inv[p % i] % p; 33 rep(i, 1, n){ 34 int k; 35 scanf("%d", &k); 36 rep(j, 1, k){ 37 int x; 38 scanf("%d", &x); 39 a[i].pb(x); 40 if(id[x] != i){ 41 sum[x]++; 42 id[x] = i; 43 } 44 } 45 } 46 int ans = 0; 47 rep(i, 1, n){ 48 rep(j, 0, SZ(a[i]) - 1){ 49 ans = (ans + 1ll * sum[a[i][j]] * inv[n] % p * inv[n] % p * inv[SZ(a[i])] % p) % p; 50 } 51 } 52 printf("%d\n", ans); 53 return 0; 54 }
-D
原文地址:https://www.cnblogs.com/nblyz2003/p/12168346.html
时间: 2024-10-28 23:49:31