HDU5742：It's All In The Mind（模拟+贪心 ）

题意：

给出n和m,表示n个数,之后会给出m个下标xi和值yi，a[xi]=yi,n个数是不下降的，且总和>0，要使得(x1+x2)/∑(xi)最大。

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分析：

尽可能使得前两个数最大，其他数尽可能小即可。

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代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, a, b)    for (int i(a); i <= (b); ++i)
int a[];
int T;
int n, m, x1,x2, y1,y2;

inline int gcd(int a, int b){return b == ? a : gcd(b, a % b);}

int main(){
    scanf("%d ", &T);
  loop:  while (T--){
        scanf("%d %d ", &n, &m);
        memset(a,, sizeof a);
        x1=;
        scanf("%d %d",&x2,&y2);
        if(x2==)
        {
            a[]=a[]=y2;x1=x2;
        }
        else
        {
            a[x2]=y2;
            if(x2<=) rep(i,,x2-) a[i]=;
            else{a[]=a[]=;rep(i,,x2-) a[i]=y2;} x1=x2;
        }
        rep(i,,m)
            {
            scanf("%d %d", &x2, &y2);
            a[x2]=y2;
            if(i<=m)
            {
                if(x1+==) rep(j,,x2-) a[j]=y2;
                else rep(j,x1+,x2-) a[j]=y2;
            }
            x1=x2;
            }
        rep(j,x1+,n) a[j]=;
        //rep(i,1,n) printf("a[%d]=%d\n",i,a[i]);
        if (n ==){ puts("1/1"); goto loop;}
        int sum =;
        rep(i,, n) sum += a[i];
        //printf("a[1]+a[2]=%d sum=%d\n",a[1]+a[2],sum);
        int g = gcd(a[]+a[], sum);
        printf("%d/%d\n", (a[]+a[])/ g, sum / g);
    }
       return;
}