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： 2024-08-22 16:40:58