标签：for 合并 read set [] space 数列 ide 技术分享
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\n",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); }
标签：for 合并 read set [] space 数列 ide 技术分享
原文地址：https://www.cnblogs.com/Kong-Ruo/p/9822453.html
