B表示完全看不懂。。就不弄了。。
E字符串先不管了。到时候系统学下字符串再处理
A
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} int src[13]; int main() { int K; scanf("%d",&K); for (int i = 0; i < 12; i++) scanf("%d",&src[i]); sort(src,src+12); int ans = 0 ,sum = 0, cas = 11; while (true) { if (sum >= K) break; if (ans == 13) break; sum += src[cas]; cas--; ans++; } if (ans == 13) puts("-1"); else printf("%d ",ans); return 0; }
C
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} #define MAXN 100005 int N; struct node { int id; int val; friend bool operator < (const node & a,const node &b) { return a.val < b.val; } }src[MAXN]; LL sum[MAXN]; node stl[MAXN],str[MAXN]; int main() { scanf("%d",&N); sum[0] = 0; for (int i = 1; i <= N; i++) { scanf("%d",&src[i].val); src[i].id = i ; sum[i] = sum[i - 1] + src[i].val; } sort(src + 1, src + 1 + N); int topl = 1 ,topr = 1; if (N % 2 == 0) { for (int i = 1; i <= N; i++) { if (i % 2 == 0) stl[topl++] =src[i]; else str[topr++] = src[i]; } printf("%d ",N / 2); for (int i = 1; i < topl; i++) printf("%d ",stl[i].id); putchar(' '); printf("%d ",N / 2); for (int i = 1; i < topr; i++) printf("%d ",str[i].id); putchar(' '); } else { for (int i = 1; i <= N; i += 2) stl[topl++] =src[i]; for (int i = 2; i <= N; i += 2) str[topr++] = src[i]; printf("%d ",N / 2 + 1); for (int i = 1; i < topl; i++ ) printf("%d ",stl[i].id); putchar(' '); printf("%d ", N / 2); for (int i = 1; i < topr; i++) printf("%d ",str[i].id); putchar(' '); } return 0; }
D
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} #define MOD 1000000007 char input[720]; struct node { char res; int id; }src[720]; int G[720]; node sta[720]; LL dp[720][720][4][4]; void calcu(int l ,int r) { if (l >= r) return ; if (l + 1 == r) { dp[l][r][0][1] = 1; dp[l][r][1][0] = 1; dp[l][r][2][0] = 1; dp[l][r][0][2] = 1; } if (G[l] == r) { calcu(l + 1,r - 1); for (int i = 0 ; i < 3; i++) for (int j = 0; j < 3; j++) { if (j != 1) dp[l][r][0][1] = (dp[l][r][0][1] + dp[l + 1][r - 1][i][j]) % MOD; if (j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + dp[l + 1][r - 1][i][j]) % MOD; if (i != 1) dp[l][r][1][0] = (dp[l][r][1][0] + dp[l + 1][r - 1][i][j]) % MOD; if (i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l + 1][r - 1][i][j]) % MOD; } return ; } else { int t = G[l]; calcu(l,t); calcu(t + 1, r); for (int i = 0 ; i < 3; i++) for (int j = 0 ; j < 3; j++) for (int m = 0; m < 3; m++) for (int n = 0 ; n < 3; n++) { if (!( (m == 1 && n == 1) || (m == 2 && n == 2))) dp[l][r][i][j] = (dp[l][r][i][j] + (dp[l][t][i][m] * dp[t + 1][r][n][j]) % MOD) % MOD; } } } int main() { while (scanf("%s",input + 1) != EOF) { int len = strlen(input + 1); int top = 1; for (int i = 1; i <= len; i++) { if (input[i] == '(') { node tmp; tmp.id = i; tmp.res = input[i]; sta[top++] = tmp; } else { if (sta[top - 1].res == '(') { G[i] = sta[top - 1].id; G[sta[top - 1].id] = i; top--; } else { node tmp; tmp.id = i; tmp.res = input[i]; sta[top++] = tmp; } } } memset(dp,0,sizeof(dp)); calcu(1,len); LL ans = 0; for (int i = 0 ; i < 3; i++) for (int j = 0 ; j < 3; j++) ans = (ans + dp[1][len][i][j]) % MOD; printf("%I64d ",ans % MOD); } return 0; }