题意:找到[a, b]的最大公约数;
思路:相同时为本身,不同时为1.
套路:碰到水题别想太多;
猜想:两个相邻数,必有一奇一偶,如果偶数有因子3或者其他,奇数可不可能有相同的呢?
枚举一些数后发现没有,出现的奇数全是素数。
1 #include <iostream> 2 #include <algorithm> 3 #include <stdlib.h> 4 #include <time.h> 5 #include <cmath> 6 #include <cstdio> 7 #include <string> 8 #include <cstring> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <set> 13 14 #define c_false ios_base::sync_with_stdio(false); cin.tie(0) 15 #define INF 0x3f3f3f3f 16 #define INFL 0x3f3f3f3f3f3f3f3f 17 #define zero_(x,y) memset(x , y , sizeof(x)) 18 #define zero(x) memset(x , 0 , sizeof(x)) 19 #define MAX(x) memset(x , 0x3f ,sizeof(x)) 20 #define swa(x,y) {LL s;s=x;x=y;y=s;} 21 using namespace std ; 22 typedef long long LL; 23 const int N = 205; 24 25 char n[N], m[N]; 26 27 int main(){ 28 //freopen("in.txt","r",stdin); 29 //freopen("out.txt","w",stdout); 30 //ios_base::sync_with_stdio(false); cin.tie(0); 31 scanf("%s%s", n, m); 32 string n1, m1; 33 n1 = n; m1 = m; 34 if(n1 == m1) printf("%s ", n); 35 else printf("1 "); 36 return 0; 37 }
题意:找到一些数,满足以下等式 :?+ ? + ?- ? = n
思路:1) 可以将+ 和 - 分开计算,+ sum1, - sum2 然后将sum1 与sum 2 分派下去,
2) (注意到数是可以重复的) 先将所有数设置为1, 然后 ans > n, - 加1,否则 + 加1;
1 #include <iostream>
2 #include <algorithm>
3 #include <stdlib.h>
4 #include <time.h>
5 #include <cmath>
6 #include <cstdio>
7 #include <string>
8 #include <cstring>
9 #include <vector>
10 #include <queue>
11 #include <stack>
12 #include <set>
13
14 #define c_false ios_base::sync_with_stdio(false); cin.tie(0)
15 #define INF 0x3f3f3f3f
16 #define INFL 0x3f3f3f3f3f3f3f3f
17 #define zero_(x,y) memset(x , y , sizeof(x))
18 #define zero(x) memset(x , 0 , sizeof(x))
19 #define MAX(x) memset(x , 0x3f ,sizeof(x))
20 #define swa(x,y) {LL s;s=x;x=y;y=s;}
21 using namespace std ;
22 typedef long long LL;
23 const int N = 205;
24
25 int num1[N], num2[N];
26 char s[2];
27
28 int main(){
29 //freopen("in.txt","r",stdin);
30 //freopen("out.txt","w",stdout);
31 //ios_base::sync_with_stdio(false); cin.tie(0);
32 zero(num1);
33 zero(num2);
34 scanf("%s", s);
35 scanf("%s", s);
36 int p = 1, q = 0, k = 1;
37 num1[1] = 1;
38 while(s[0] != '='){
39 if(s[0] == '+'){
40 num1[++k] = 1;
41 p++;
42 }
43 if(s[0] == '-'){
44 num1[++k] = -1;
45 q++;
46 }
47 scanf("%s", s);
48 scanf("%s", s);
49 }
50 int n;
51 scanf("%d", &n);
52 for(int i = 1; i <= k; i++){
53 num2[i] = 1;
54 }
55 int sum = p - q;
56 for(int i = 1; i <= k; i++){
57 while(sum < n && num1[i] == 1 && num2[i] < n){
58 num2[i]++, sum++;
59 }
60 while(sum > n && num1[i] == -1 && num2[i] < n){
61 num2[i]++, sum--;
62 }
63 }
64 if(sum != n){
65 printf("Impossible
");
66 return 0;
67 }
68 printf("Possible
");
69 printf("%d ", num2[1]);
70 for(int i = 2; i <= k; i++){
71 if(num1[i] == 1) printf("+ ");
72 else if(num1[i] == -1) printf("- ");
73 printf("%d ", num2[i]);
74 }
75 printf("= %d
", n);
76 return 0;
77 }
C:
题意:求尾号代表的年份
思路:贪心法, 主要是如何通过后缀找到原年份。
区间的长度为后缀字符的个数^10
可以先预处理区间长度再找后缀字符的位置;
1 #include <iostream> 2 #include <algorithm> 3 #include <stdlib.h> 4 #include <time.h> 5 #include <cmath> 6 #include <cstdio> 7 #include <string> 8 #include <cstring> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <set> 13 14 #define c_false ios_base::sync_with_stdio(false); cin.tie(0) 15 #define INF 0x3f3f3f3f 16 #define INFL 0x3f3f3f3f3f3f3f3f 17 #define zero_(x,y) memset(x , y , sizeof(x)) 18 #define zero(x) memset(x , 0 , sizeof(x)) 19 #define MAX(x) memset(x , 0x3f ,sizeof(x)) 20 #define swa(x,y) {LL s;s=x;x=y;y=s;} 21 using namespace std; 22 typedef long long LL; 23 const int N = 205; 24 25 LL a[30], a1[30]; 26 char b[20]; 27 28 int Solve(){ 29 int len = strlen(b); 30 LL num = 0; 31 for(int i = 4; i < len; i++){ 32 num *= 10; 33 num += b[i] - '0'; 34 } 35 for(int i = 0; i < 300; i++){ 36 int number = num + (LL)i * a1[len-4]; 37 if(number >= a[len-5] && number < a[len - 4]){ 38 printf("%d ", number); 39 return 0; 40 } 41 } 42 return 0; 43 } 44 45 int main(){ 46 //freopen("in.txt","r",stdin); 47 //freopen("out.txt","w",stdout); 48 //ios_base::sync_with_stdio(false); cin.tie(0); 49 int n, k = 1; 50 a[0] = 1989; 51 a1[0] = 1; 52 //设置首位置 53 for(int i = 1; i < 15; i++){ 54 k *= 10; 55 a[i] = a[i-1] + k; 56 a1[i] = k; 57 } 58 scanf("%d", &n); 59 while(n--){ 60 scanf("%s", b); 61 //printf("%s ", b); 62 Solve(); 63 } 64 return 0; 65 }
D:
题意:将一些点翻转过来,得到所有边都相同颜色的图
思路:暴力美学,dfs搜索