130827解题报告

来源：互联网发布：js求二维数组的最大值编辑：程序博客网时间：2024/04/30 13:13

A，B，J三题较为水，算是一眼题了

C. Pen Counts

题意：用1--n之间的数组成符合题意的三角形（每个数只能用一次），求三角形的个数。

经过统计，如果三条边的值完全不同，三角形个数+2，否则三角形个数+1，然后用过的三条边，就不能再用了。所以直接暴力找，中间弄点剪枝就能过了。

#include <cstdio>#include <iostream>#include <cmath>#include <cstring>using namespace std;int main() {    int T,ca,n;    scanf("%d",&T);    while(T--) {        scanf("%d%d",&ca,&n);        int sum = 0;        int len = n / 2;        for(int i=len; i>=1; i--) {            for(int j=i; j>=1; j--) {                int k = n - i - j;                if(k > j) break ;                if(k + j > i) {                    if(i != j && j != k && i != k) sum += 2;                    else sum ++;                }            }        }        printf("%d %d\n",ca,sum);    }    return 0;}

D. Maximum Random Walk

一道概率DP题，用dp[i][j][k]表示走了i步，当前走在j点，过程中已经走过的最靠右边的点为k的概率..........相当暴力，二维的做法表示不懂

#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100005#define INF 0x7FFFFFFF#define REP(i,s,t) for(int i=(s);i<=(t);++i)#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define L(x) x<<1#define R(x) x<<1|1# define eps 1e-5//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂using namespace std;double ll,rr,stop;double dp[2][2111][2111];int n,ca;void solve() {    memset(dp,0,sizeof(dp));    dp[0][1000][1000] = 1;    int cur = 0;    for(int i=1; i<=n; i++) {        for(int j=1000-i; j<=1000+i; j++) {            int k = max(j,1000);            for(; k<=i+1000; k++) {                if(k > j) dp[1-cur][j][k] = dp[cur][j][k] * stop + dp[cur][j-1][k] * rr + dp[cur][j+1][k] * ll;                else dp[1-cur][j][k] = dp[cur][j][k] * stop + (dp[cur][j-1][k] + dp[cur][j-1][j-1])* rr;            }        }        cur = 1 - cur;    }    double sum = 0;    for(int j=1000-n; j<=1000+n; j++) {        int k = max(j,1000);        for(; k<=1000+n; k++) {            sum += dp[cur][j][k] * (k - 1000);        }    }    printf("%d %.4f\n",ca,sum);}int main() {    int T;    cin >> T;    while(T--) {        scanf("%d%d",&ca,&n);        scanf("%lf%lf",&ll,&rr);        stop = 1 - ll - rr;        solve();    }    return 0;}

F. The King's Ups and Downs

题意：给定了1---n的数字，现在要把他们排列：高低高低的排列，或者低高低高的排列...求出所有的排列数

表示数学能力太差了，只能写暴力的状态压缩DP..........dp[i][j][k]表示排到第i个位置，状态为j，第i-1个位置高于或者低于第i个位置时，总的排列数。状压完打表即可。

#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <set>#include <queue>#include <stack>#include <climits>//形如INT_MAX一类的#define MAX 100005#define INF 0x7FFFFFFF#define REP(i,s,t) for(int i=(s);i<=(t);++i)#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define L(x) x<<1#define R(x) x<<1|1# define eps 1e-5//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂using namespace std;int n;//long long dp[22][1<< 20][2];////long long solve() {//    memset(dp,0,sizeof(dp));//    for(int i=1; i<=n; i++) {//        dp[i][1<<(i-1)][0] = 1;//        dp[i][1<<(i-1)][1] = 1;//    }//    int total = 1 << n;//    for(int j=1; j<total; j++) {//        for(int i=1; i<=n; i++) {//            for(int k=1; k<i; k++) {//                if((j & (1<<(i-1))) && (j & (1 << (k-1)))) {//                    dp[i][j][0] += dp[k][j ^ (1 << (i-1))][1];//                }//            }//            for(int k=i+1; k<=n; k++) {//                if((j & (1 << (i-1))) && (j & (1 << (k-1)))) {//                    dp[i][j][1] += dp[k][j ^ (1 << (i-1))][0];//                }//            }//        }//    }//    long long sum = 0;//    for(int i=1; i<=n; i++) {//        sum +=  dp[i][(1<<n) - 1][0] + dp[i][(1<<n) - 1][1];//    }//    return sum;//}long long table[] = {0,1,2,4,10,32,122,544,2770,15872,101042,707584,                    5405530,44736512,398721962,3807514624LL,38783024290LL,                    419730685952LL,4809759350882LL,58177770225664LL,                    740742376475050LL};int main() {    int T,ca;    cin >> T;    while(T--) {        scanf("%d%d",&ca,&n);        printf("%d %lld\n",ca,table[n]);    }    return 0;}