ACdream P1174 Sum

Sum

Time Limit: 6000/3000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

给出N,a[1]... a[N]，还有M,b[1]... b[M]
long long ans = 0;
for(int i = 1; i <= N; i ++)
    for(int j = 1; j <= M; j ++)
        ans += abs(a[i] - b[j]) * (i - j);

Input

多组数据，每组数据

第一行N,M(1 <= N,M <= 50000)

第二行N个数字，a[1].. a[N]

第三行M个数字，b[1]..b[M]

(1 <= a[i],b[i] <= 10000)

Output

每组数据一行，ans

Sample Input

4 41 2 3 45 6 7 8

Sample Output

-40

Hint

you may be TLE if 10000 * 10000 per case

Source

classical

Manager

scaucontest

刚做的一道水题，趁热写一发题解。

这题先考虑没有ABS的情况(ai-bj)*(i-j)把这之类所有式子都写出，然后合并同类项，可得a数组的项数符合-m(m-1)/2为首项，差为m的等差数列，b数组同理。

再考虑有了绝对值怎么办，有了绝对值无非是一些小于bj的ai项数（bj类似）出现了变动，故先排序，这样就可以清楚求出那些ai碰到了哪些bj，发生相应的项数变动。

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<ctime>#include<string>#include<cstring>#include<algorithm>#include<queue>#include<stack> #include<vector>#include<cmath>#define rep(i,n) for(i=1;i<=n;i++)#define MM(a,t) memset(a,t,sizeof(a))#define INF 1e9typedef long long ll;#define mod 1000000007using namespace std;ll ans;struct node{  ll  val,index;  char zl;}a[100002];ll cnta[50002],cntb[50002];ll n,m; void setup(){<span style="white-space:pre"></span>ll i,j;<span style="white-space:pre"></span>ll tmp;<span style="white-space:pre"></span><span style="white-space:pre"></span>cnta[1]=-(m*(m-1)/2);<span style="white-space:pre"></span>for(i=2;i<=n;i++)<span style="white-space:pre"></span>  cnta[i]=cnta[i-1]+m;    cntb[1]=-(n*(n-1)/2);    for(i=2;i<=m;i++)      cntb[i]=cntb[i-1]+n;}bool cmp1(node a1,node a2){<span style="white-space:pre"></span>return a1.val<a2.val;}ll ReadInt()//输入挂 {    char ch = getchar();    ll data = 0;    while (ch < '0' || ch > '9')        ch = getchar();    do    {        data = data*10 + ch-'0';        ch = getchar();    } while (ch >= '0' && ch <= '9');    return data;}int main(){<span style="white-space:pre"></span>ll i,j;<span style="white-space:pre"></span><span style="white-space:pre"></span>while(scanf("%lld%lld",&n,&m)!=EOF){<span style="white-space:pre"></span>setup(); ans=0;<span style="white-space:pre"></span>rep(i,n){<span style="white-space:pre"></span>ll tmp;<span style="white-space:pre"></span>tmp=ReadInt();<span style="white-space:pre"></span>a[i].zl='a';<span style="white-space:pre"></span>a[i].val=tmp;<span style="white-space:pre"></span>a[i].index=i;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>for(i=n+1;i<=n+m;i++){<span style="white-space:pre"></span>ll tmp;<span style="white-space:pre"></span>tmp=ReadInt();<span style="white-space:pre"></span>a[i].zl='b';<span style="white-space:pre"></span>a[i].val=tmp;<span style="white-space:pre"></span>a[i].index=i-n;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>sort(a+1,a+n+m+1,cmp1);<span style="white-space:pre"></span>ll s1=0,sum1=0;<span style="white-space:pre"></span>rep(i,n+m)<span style="white-space:pre"></span>if(a[i].zl=='a'){<span style="white-space:pre"></span>  s1+=1; sum1+=a[i].index;<span style="white-space:pre"></span><span style="white-space:pre"></span>}<span style="white-space:pre"></span>else cntb[a[i].index]+=2*(sum1-s1*a[i].index);<span style="white-space:pre"></span>s1=0; sum1=0;<span style="white-space:pre"></span>for(i=n+m;i>=1;i--)<span style="white-space:pre"></span>if(a[i].zl=='b'){<span style="white-space:pre"></span>s1+=1; sum1+=a[i].index;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>else cnta[a[i].index]-=2*(s1*a[i].index-sum1);<span style="white-space:pre"></span>rep(i,n+m)<span style="white-space:pre"></span>if(a[i].zl=='a') ans+=cnta[a[i].index]*a[i].val;<span style="white-space:pre"></span>else             ans+=cntb[a[i].index]*a[i].val;<span style="white-space:pre"></span>printf("%lld\n",ans);<span style="white-space:pre"></span>}<span style="white-space:pre"></span><span style="white-space:pre"></span>return 0;}

方法一般，想上进的同学可以去探求更快的方法。出题者卡了个常数来卡我这种做法，但是我机智地用了输入挂破解了，求管理员继续卡啊！你要再卡常数这道题就没意思了，题目出的不够好请别用卡常数这种低级无聊手段让我们臣服。