The 2016 ACM-ICPC Asia Shenyang Regional Contest

A. Thickest Burger

大数 × 2 + 小数

#include <cstdio>
#include <algorithm>
using namespace std;

int T;
int A,B;
int main()
{
    scanf("%d",&T);
    for(int t=1; t<=T; t++)
    {
        scanf("%d%d",&A,&B);
        if(A<B) swap(A,B);
        printf("%d
",A*2+B);
    }
    return 0;
}

　　

B. Relative atomic mass

给定一个分子式，只包含 H C O 三种，求相对分子质量。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=15;

char s[maxn];
int T, ans = 0;
int main()
{
    scanf("%d",&T);
    for(int t=1; t<=T; t++)
    {
        ans=0;
        scanf("%s",s);
        int n=strlen(s);
        for(int i=0; i<n; i++)
        {
            if(s[i]=='H') ans++;
            if(s[i]=='C') ans+=12;
            if(s[i]=='O') ans+=16;
        }
        printf("%d
",ans);
    }
    return 0;
}

　　

C. Recursive sequence

矩阵快速幂

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const LL MOD = 2147493647;
int a, b;
LL C[7][7];

void mut(LL A[][7], LL B[][7]) {
    memset(C, 0, sizeof(C));
    for(int i = 0; i < 7; ++i)
        for(int j = 0; j < 7; ++j)
            for(int k = 0; k < 7; ++k)
                C[i][j] = ( C[i][j] + A[i][k] * B[k][j] ) % MOD;
    memcpy(A, C, sizeof(C));
}

LL qpow(int n) {
    LL aa[7][7] = {{1,2,1,0,0,0,0},{1,0,0,0,0,0,0},{0,0,1,4,6,4,1},{0,0,0,1,3,3,1},{0,0,0,0,1,2,1},{0,0,0,0,0,1,1},{0,0,0,0,0,0,1}};
    LL ans[7][7] = {{1,0,0,0,0,0,0},{0,1,0,0,0,0,0},{0,0,1,0,0,0,0},{0,0,0,1,0,0,0},{0,0,0,0,1,0,0},{0,0,0,0,0,1,0},{0,0,0,0,0,0,1}};
    while(n) {
        if(n&1) mut(ans, aa);
        mut(aa,aa);
        n>>=1;
    }
    LL res = ans[0][0] * b % MOD + ans[0][1] * a % MOD + ans[0][2] * 3 * 3 * 3 * 3 % MOD;
       res = res + ans[0][3] * 3 * 3 * 3 % MOD + ans[0][4] * 3 * 3 % MOD + ans[0][5] * 3 % MOD;
       res = ( res + ans[0][6] ) % MOD;
    return res;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d%d%d", &n, &a, &b);
        if(n == 1) {
            printf("%d
",a);
            continue;
        }
        LL ans = qpow(n-2);
        printf("%lld
", ans);
    }
    return 0;
}

　　

D. Winning an Auction

博弈

E. Counting Cliques

爆搜。vector[i] 记录与 i 有边且编号大于 i 的点。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 100 + 10;
const int maxm = 1000 + 100;

int n, m, k;
vector<int> v[maxn];
int from[maxm], to[maxm];
int G[maxn][maxn];
int d[maxn], node[maxn];
int tot, ans;

void DFS(int x, int start)
{
    if (tot == k) { ++ans; return; }

    int sz = v[x].size();
    for (int i = start; i < sz; i++)
    {
        int flag = 0;
        for (int j = 2; j <= tot; j++)
        if (!G[ node[j] ][ v[x][i] ]) { flag = 1; break; }

        if (flag) continue;

        node[++tot] = v[x][i], DFS(x, i+1), --tot;
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    for (int ca = 1; ca <= t; ca++)
    {
        memset(d, 0, sizeof(d));
        for (int i = 1; i <= n; i++)
        {
            for (int j = i+1; j <= n; j++) G[i][j] = G[j][i] = 0;
            v[i].clear();
        }

        scanf("%d%d%d", &n, &m, &k);
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d", &from[i], &to[i]);
            d[ from[i] ]++, d[ to[i] ]++;
        }

        for (int i = 1; i <= m; i++)
            if (d[ from[i] ] >= k-1 && d[ to[i] ] >= k-1)
            {
                if (from[i] < to[i]) v[ from[i] ].push_back(to[i]);
                    else v[ to[i] ].push_back(from[i]);
                G[ from[i] ][ to[i] ] = G[ to[i] ][ from[i] ] = 1;
            }

        ans = 0;
        for (int i = 1; i <= n; i++)
        {
            tot = 1, node[1] = i;
            DFS(i, 0);
        }

        printf("%d
", ans);
    }
}

　　

F. Similar Rotations

G. Do not pour out

H. Guessing the Dice Roll

I. The Elder

J. Query on a graph

K. New Signal Decomposition

L. A Random Turn Connection Game

M. Subsequence