A：贪心，遍历每次维护一个最便宜的价格，假如当前价格不如此前价格，就用此前价格购买当前数量的肉，每次更新最便宜的价格。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 typedef struct Node {
23 int a;
24 int p;
25 int day;
26 }Node;
27 const int maxn = 100010;
28 int n;
29 Node orz[maxn];
30
31
32 int main() {
33 // freopen("in", "r", stdin);
34 while(~scanf("%d", &n)) {
35 int minn = 0x7f7f7f;
36 int midx;
37 int ans = 0;
38 for(int i = 1; i <= n; i++) {
39 scanf("%d %d", &orz[i].a, &orz[i].p);
40 if(minn > orz[i].p) {
41 minn = orz[i].p;
42 ans += minn * orz[i].a;
43 }
44 else {
45 ans += minn * orz[i].a;
46 }
47 }
48 cout << ans << endl;
49 }
50 return 0;
51 }
A
B：根据一个性质2^k=2*2^(k-1)，每次计数不同幂的个数，假如有偶数个就向下一个数进位并划归为下一个数的个数(k/2个），直到k是奇数的时候，这时候举一次再更新。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 typedef long long ll;
23 const int maxn = 1100010;
24 int n;
25 ll dp[maxn];
26
27 int main() {
28 // freopen("in", "r", stdin);
29 int w;
30 while(~scanf("%d", &n)) {
31 memset(dp, 0, sizeof(dp));
32 for(int i = 1; i <= n; i++) {
33 scanf("%d", &w);
34 dp[w]++;
35 }
36 ll ans = 0;
37 for(int i = 0; i < maxn; i++) {
38 dp[i+1] += dp[i] / 2;
39 ans += dp[i] % 2;
40 }
41 printf("%I64d\n", ans);
42 }
43 return 0;
44 }
B
C：利用fibonacci通项公式，先看对数的性质,loga(b^c)=c*loga(b),loga(b*c)=loga(b)+loga(c);假设给出一个数10234432,
那么log10(10234432)=log10(1.0234432*10^7)【用科学记数法表示这个数】=log10(1.0234432)+7;
log10(1.0234432)就是log10(10234432)的小数部分.
log10(1.0234432)=0.010063744(取对数所产生的数一定是个小数)
再取一次幂:10^0.010063744=1.023443198
取前4位，只需要将这个结果乘1000就可以了。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 typedef long long LL;
23 int f[21];
24 int n;
25
26 void init() {
27 f[0] = 0;
28 f[1] = 1;
29 for(int i = 2; i < 21; i++) {
30 f[i] = f[i-1] + f[i-2];
31 }
32 }
33
34 void solve() {
35 if(n < 21) {
36 printf("%d\n", f[n]);
37 }
38 else {
39 int answer;
40 double ans = -0.5 * log10(5.0) + n * log10((1+sqrt(5))/2);
41 ans -= floor(ans);
42 ans = pow(10, ans);
43 answer = int(ans * 1000);
44 printf("%d\n", answer);
45 }
46 }
47
48 int main() {
49 // freopen("in", "r", stdin);
50 init();
51 while(~scanf("%d", &n)) {
52 solve();
53 }
54 return 0;
55 }
C
D：找出a数组里第k小的，再找出b数组里第m大的，假如a数组里k小的数都比b数组里m大的数小，那就满足条件。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 const int maxn = 100010;
23 int na, nb;
24 int k, m;
25 int a[maxn], b[maxn];
26
27 int main() {
28 // freopen("in", "r", stdin);
29 while(~scanf("%d %d", &na, &nb)) {
30 scanf("%d %d", &k, &m);
31 for(int i = 1; i <= na; i++) scanf("%d", &a[i]);
32 for(int i = 1; i <= nb; i++) scanf("%d", &b[i]);
33 sort(a+1, a+na+1);
34 sort(b+1, b+nb+1);
35 if(a[k] < b[nb-m+1]) puts("YES");
36 else puts("NO");
37 }
38
39 return 0;
40 }
D
E:某人有n个朋友，这n个朋友有钱数m和关系s两个属性。问如何选择朋友，使得这些朋友之间s最大差距小于d并且钱数是最多。
可以用滑动窗口，将m从小到大，s从大到小排列，这时在一个队列里维护队首和队尾，假如队首和队尾的s差距≥d时，就把队尾扔掉队首入队否则就仅队首入队。此时更新一下当前最大值。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 typedef long long ll;
23 typedef struct Node {
24 int m;
25 int s;
26 }Node;
27 const int maxn = 100010;
28 int n, d;
29 Node f[maxn];
30
31 bool cmp(Node a, Node b) {
32 if(a.m == b.m) return a.s > b.s;
33 return a.m < b.m;
34 }
35
36 int main() {
37 // freopen("in", "r", stdin);
38 while(~scanf("%d %d", &n, &d)) {
39 for(int i = 1; i <= n; i++) {
40 scanf("%d %d", &f[i].m, &f[i].s);
41 }
42 sort(f+1, f+n+1, cmp);
43 ll curans = 0;
44 ll ans = 0;
45 int front = 1;
46 int tail = 1;
47 while(1) {
48 if(front > n || tail > n) break;
49 if(f[tail].m - f[front].m >= d)
50 curans -= f[front++].s;
51 else curans += f[tail++].s;
52 ans = max(ans, curans);
53 }
54 printf("%I64d\n", ans);
55 }
56 return 0;
57 }
E
F:先排序，知道第1个肯定和第n个离得最远，而和第2个离得最近。同理第n个和第1个离得最远，和第n-1个离得最近。固定这两个，接下来在中间找i，最远的话和1和n比，最近的话和i-1和i+1比。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 const int maxn = 1555555;
23 int n;
24 int x[maxn];
25
26 int main() {
27 // freopen("in", "r", stdin);
28 while(~scanf("%d", &n)) {
29 for(int i = 1; i <= n; i++) {
30 scanf("%d", &x[i]);
31 }
32 sort(x+1,x+n+1);
33 printf("%d %d\n", abs(x[1]-x[2]), abs(x[1]-x[n]));
34 for(int i = 2; i < n; i++) {
35 printf("%d %d\n", abs(x[i]-x[i-1]<abs(x[i]-x[i+1]))?abs(x[i]-x[i-1]):abs(x[i]-x[i+1]),
36 (abs(x[i]-x[n])>abs(x[i]-x[1]))?abs(x[i]-x[n]):abs(x[i]-x[1]));
37 }
38 printf("%d %d\n", abs(x[n]-x[n-1]), abs(x[1]-x[n]));
39 }
40 return 0;
41 }
F
G:按照从左到右的顺序，找一个子串。使得这个子串看成是上下左右移动步骤的时候可以走回远点。算算每一个子串里上下左右的个数就行了，假如上的次数等于下的次数，左的次数等于右的次数就说明可以回到原点。
1 #include <algorithm>
2 #include <iostream>
3 #include <iomanip>
4 #include <cstring>
5 #include <climits>
6 #include <complex>
7 #include <fstream>
8 #include <cassert>
9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19
20 using namespace std;
21
22 const int maxn = 555555;
23 int n;
24 char str[maxn];
25
26 int main() {
27 // freopen("in", "r", stdin);
28 while(~scanf("%d", &n)) {
29 scanf("%s", str);
30 int ans = 0;
31 int u, d, l, r;
32 for(int i = 0; str[i]; i++) {
33 u = d = l = r = 0;
34 for(int j = i; str[j]; j++) {
35 if(str[j] == ‘U‘) u++;
36 if(str[j] == ‘D‘) d++;
37 if(str[j] == ‘L‘) l++;
38 if(str[j] == ‘R‘) r++;
39 if(l == r && u == d) {
40 for(int k = i; k <= j; k++) {
41 printf("%c", str[k]);
42 }
43 printf("\n");
44 ans++;
45 }
46 }
47 }
48 printf("%d\n", ans);
49 }
50 return 0;
51 }
G
时间： 2024-11-20 02:28:25