Loading

Xeon 第一次训练赛 苏州大学ICPC集训队新生赛第二场(同步赛) [Cloned]

A.给出一个字符串,求出连续的权值递增和,断开以后权值重新计数,水题

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
//const int maxn = 1e6 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

char s[85];

int main() {
    int T;
    int ans;
    int cnt;
    scanf("%d", &T);
    getchar();
    while (T--) {
        ans = 0;
        cnt = 1;
        scanf("%s", s);
        for (int i = 0; i < strlen(s); i++) {
            if (s[i] != 'X') ans += cnt, cnt++;
            else cnt = 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}
View Code

B.给出四面体ABCD,起点由D开始走n步回到D,问有多少种不同的走法(中间可以再次经过D)

考虑DP枚举,复杂度O(n)

#include<cstdio>
using namespace std;
const int maxn = 1e7+5;
const int mod = 1e9+7;
int dp[maxn][4];
int n;
int main(){
    scanf("%d",&n);
    dp[1][0] = 0;
    dp[1][1] = dp[1][2] = dp[1][3] = 1;
    for (int i = 1; i <= n; i++){
        for (int j = 0; j < 4; j++){
            for (int k = 0; k < 4; k++){
                if (j == k) continue;
                dp[i][j] += dp[i-1][k];
                dp[i][j] %= mod;                    
            }
        }
    }
    printf("%d",dp[n][0]);
    return 0;
}
View Code

C.给出数字a,b,求a的重新排列以后小于b的最大值

思维题,若|a|<|b|,显然只需将a降序排列即可,否则 方法是将a升序排列,每次确定最高位,最终结果就是最优解

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
const int maxn = 1e9 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

string a;
string b;
string t;

int main() {
    cin >> a >> b;
    int len1 = a.length(), len2 = b.length();
    sort(a.begin(), a.end());
    if (len1 < len2) reverse(a.begin(), a.end());
    else {
        int l, r;
        for (l = 0; l < len1; l++) {
            r = len1 - 1;
            t = a;
            while (r > l) {
                swap(a[l], a[r--]);
                sort(a.begin()+l+1, a.end());
                if (a > b) a = t;
                else break;
            }
        }

    }
    printf("%s", a.c_str());
    return 0;
}
View Code

G.从起点出发到目标点x,步数从1开始每次严格增加1,问最短步次可达x

仍然是思维题 首先显然的是x是负数无需考虑(对称性), 接下来考虑x在数轴右侧

首先考虑x无限大的情况,那显然前期的任何一步都是浪费的(物理思维?),事实上也是如此,往左走的步数显然是用于微调的

考虑到一直往左走第一次超越x,若此时del是偶数,那么显然可以在之前的第del/2步往左走,效果相当于往左走了del步,正好可达最优解

若del是奇数,那只需至多再走1,2步就能到偶数的情况

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
const int maxn = 1e9 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

int ans[100005];

int main() {
    int i;
    for(i=1;;i++){
        ans[i]=i*(i+1)/2;
        if(ans[i]>maxn) break;
    }
    
    int x;
    scanf("%d",&x);
    if(!x) {
        printf("0");
        return 0;
    }
    if(x<0) x=-x;
    int p=lower_bound(ans,ans+i,x)-ans;
    int ret=ans[p]-x;
    while(ret&1) p++,ret+=p;
    int Ans=p; 
    printf("%d",Ans);
    return 0;
}
View Code

F.水题, 给定n个矩形,求一点的坐标,该点满足被k个矩形包含

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
const int maxn = 1e9 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

int a[55];

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k > n) {
        printf("-1"); return 0;
    }
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    sort(a, a + n);
    printf("%d 0", a[n - k ]);
    return 0;
}
View Code

 D.数学+构造

 考虑到题目给的提示“不超过n+1‘,说明答案应该和n有关,策略是先让每个数变成大数,然后依次取模,只要取模数是递减的,就能保证是递增的

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
//const int maxn = 1e9 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

int maxn = 1e5+5000;
int a[2005];

int main() {
    int n;
    scanf("%d", &n);

    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        a[i] += maxn;
    }
    printf("%d\n", n + 1);
    printf("1 %d %d\n", n, maxn);
    for (int i = 1; i <= n; i++) {
        printf("2 %d %d\n", i, a[i] - i);
    }
    return 0;
}
View Code

 

  

posted @ 2020-02-28 13:41  MQFLLY  阅读(175)  评论(0编辑  收藏  举报