noip模拟赛

T1 飞行时间

对于一个可能跨时区的航班，给定来回程的起降时间。假设飞机来回飞行时间相同，求飞机的飞行时间。

sol：

很明显答案是(过去落地 - 过去起飞 + 回来落地 - 回来起飞) / 2

时间转换要仔细算一下不要跟某省队dalao一样写挂就可以了

T2 二阶和

区间修改，求区间区间和的和

sol：

维护一阶前缀和数列，修改就是区间加一个等差数列

只要维护每个地方的首项和公差即可

T3 合人问题

一个环，每个点有一个权值，合并两个数的代价是他们权值差的绝对值，然后会生成一个新数，权值为原来在右边那个数的权值

求把所有数合并起来的最小代价

sol：

这种傻题直接做吧

区间dp

#include<bits/stdc++.h>
#define LL long long
using namespace std;
inline int read()
{
    int x = 0,f = 1;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f;
    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0';
    return x * f;
}
int T;
char ch1[50],ch2[50],ch3[50],ch4[50],tch[50];
int solve(char ch[])
{
    int len = strlen(ch + 1);
    int pos1 = -1,pos2 = -1,plu = 0,pp = len;
    for(int i=1;i<=len;i++)
    {
        if(ch[i] == ':' && pos1 == -1)pos1 = i;
        if(ch[i] == ':' && pos1 != -1)pos2 = i;
        if(ch[i] == '(')plu = ch[i + 2] - '0',pp = i - 1;
    }
    int sum = 0,tmp = 0;sum += plu * 24 * 60 * 60;
    //cout<<plu<<endl;
    for(int i=1;i<pos1;i++)tmp = tmp * 10 + ch[i] - '0';
    sum += tmp * 60 * 60;tmp = 0;
    for(int i=pos1+1;i<pos2;i++)tmp = tmp * 10 + ch[i] - '0';
    sum += tmp * 60;tmp = 0;
    for(int i=pos2+1;i<=pp;i++)tmp = tmp * 10 + ch[i] - '0';
    sum += tmp;
    return sum;
}
int main()
{
    freopen("timezone.in","r",stdin);
    freopen("timezone.out","w",stdout);
    T = read();
    int flg = 0;
    while(T--)
    {
        if(flg){int len = strlen(tch + 1);for(int i=1;i<=len;i++)ch1[i] = tch[i];flg = 0;}
        else scanf("%s",ch1 + 1);
        scanf("%s",ch2 + 1);
        scanf("%s",tch + 1);
        if(strlen(tch + 1) == 4)
        {
            int len = strlen(ch2 + 1);
            for(int i=len+1;i<=len+4;i++)ch2[i] = tch[i - len];
            scanf("%s",ch3 + 1);
        }
        else
        {
            int len = strlen(tch + 1);
            for(int i=1;i<=len;i++)ch3[i] = tch[i];
        }
        scanf("%s",ch4 + 1);
        scanf("%s",tch + 1);
        if(strlen(tch + 1) == 4)
        {
            int len = strlen(ch4 + 1);
            for(int i=len+1;i<=len+4;i++)ch4[i] = tch[i - len];
            //scanf("%s",ch3 + 1);
        }
        else flg = 1;
        int val1 = solve(ch1),val2 = solve(ch2),val3 = solve(ch3),val4 = solve(ch4);
        int fval = -(val1 + val3 - val2 - val4) / 2;
        int a1 = fval / 60 / 60,a2 = (fval - 60 * 60 * a1) / 60,a3 = fval - 60 * 60 * a1 - 60 * a2;
        printf("%02d:%02d:%02d
",a1,a2,a3);
    }
}

#include<bits/stdc++.h>
#define inf 2139062143
#define ll long long
#define maxn 210
using namespace std;
inline int read()
{
    int x = 0,f = 1;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f;
    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0';
    return x * f;
}
int n,dp[maxn][maxn],a[maxn],b[maxn],ans=2147483233;
int main()
{
    freopen("merge.in","r",stdin);
    freopen("merge.out","w",stdout);
    n=read();memset(dp,63,sizeof(dp));
    for(int i=1;i<=n;i++) read(),b[i]=b[n+i]=read(),dp[i][i]=dp[n+i][n+i]=0;
    for(int j=1;j<=2*n;j++)for(int i=1;i+j<=2*n;i++)for(int k=i;k<i+j;k++) 
                dp[i][i+j]=min(dp[i][i+j],dp[i][k]+dp[k+1][i+j]+abs(b[i+j]-b[k]));
    for(int i=1;i<=n;i++) ans=min(ans,dp[i][i+n-1]);
    printf("%d",ans);
}