18.03.03测试

测试...挺坑的...

背包的地方想到了化为多重背包进行优化,然而写不出来...

中国剩余定理还是细节有问题,取模乘积还是不熟练

总之要多做题加强.

T1 AC 01背包数据挺水的... 板子过了

代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define rep(i,a,n) for(int i = a;i <= n;i++)
using namespace std;
typedef long long LL;
int read() {
    int ans = 0,fu = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-') fu = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        ans = ans * 10 + ch - '0';
        ch = getchar();
    }
    return ans * fu;
}

//head
const int N = 10010;
int w[N],val[N],dp[N];

int main() {
    freopen("medic.in","r",stdin);
    freopen("medic.out","w",stdout);
    int maxx = read();
    int n = read();
    rep(i,1,n) {
        w[i] = read();
        val[i] = read();
    }
    rep(i,1,n)
    for(int j = maxx; j >= w[i]; j--)
        dp[j] = max(dp[j],dp[j - w[i]] + val[i]);
    printf("%d\n",dp[maxx]);
    return 0;
}

 

T2 40% 写的格式和顺序有问题

改完标程

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
LL a[15],b[15],m[15],res[15];
LL read()
{
    LL ans = 0,fu = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') fu = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        ans = ans * 10 + ch - '0';
        ch = getchar();
    }
    return ans * fu;
}
//head 
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(!b)
    {
        x = 1,y = 0;
        return a;
    }
    LL d = exgcd(b,a % b,y,x);
    y -= x * (a / b);
    return d;
}
int main()
{
    freopen("cao.in","r",stdin);
    freopen("cao.out","w",stdout); 
    int n = read();
    LL mod = 1;
    for(int i = 1;i <= n;i++)
    {
        a[i] = read();
        mod *= a[i];
        b[i] = read();
    }
    for(int i = 1;i <= n;i++)
    {
        m[i] = mod / a[i];
        LL x,y;
        exgcd(m[i],a[i],x,y);
        //保证x*m[i]在%a[i]时为1,%a[j]时为0(j!=i) 
        x = (x + a[i]) % a[i];//x>0
        res[i] = x * m[i] * b[i];//res[i] % a[i] == b[i],res[i] % b[j] == 0(j != i);
    }
    LL ans = 0;
    for(int i = 1;i <= n;i++) ans += res[i];
    ans %= mod;
    printf("%lld\n",ans);
    return 0;
}
/*
s:3
3 1
5 1
7 2
p: 16
*/

T3 40% 数据范围扩大后可以用多重背包变nlgn

AC标程

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
using namespace std;
typedef long long LL;

int read()
{
    int ans = 0,fu = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') fu = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        ans = ans * 10 + ch - '0';
        ch = getchar();
    }
    return ans * fu;
}
//head***

const int N = 100005;
int dp[N],num[101];

void OIbag(int cst,int val,int n)//01背包封装成函数
{
    per(i,n,cst)
        dp[i] = max(dp[i],dp[i - cst] + val);
}
void cbag(int cst,int val,int n)//完全背包封装成函数
{
    rep(i,cst,n)
        dp[i] = max(dp[i],dp[i - cst] + val);
}


int mtbag(int m)//多重背包
{
    rep(i,0,100)//遍历每种物品
    {
        int c = i / 10;
        int v = i % 10;
        if(num[i] * v >= m)
            cbag(c,v,m);
            //如果全装进去已经超了重量,那么就用完全背包去套
        else {
            int k = 1;
            //取得光的话，去遍历每种取法
            //优化:这里用到是二进制思想，把复杂度从num[i]变成了lognum[i] 
            while(k < num[i]) {
                OIbag(k * c,k * v,m);
                num[i] -= k;
                k *= 2;
            }
            OIbag(num[i] * c,num[i] * v,m);//rest
        }
    }
    return dp[m];
}


int main()
{
    freopen("medic2.in","r",stdin);
    freopen("medic2.out","w",stdout);
    int n = read();
    int maxx = read();
    //优化：由于t和v排列最多101种,化为多重背包 
    rep(i,1,n)
    {
           int dx = read();
           int dy = read();
           num[dx * 10 + dy]++;
    }
    printf("%d\n",mtbag(maxx));
    return 0;
}

By ProphetB