二分答案的边界问题还是要注意

double挨着，int+1-1,

此题用到long long，所以初始化ans要足够大，前缀和优化

依然根据check答案大小左右mid，虽然有s,但是有了+1-1加持所以能够自动推出

#include
     
      #define int long long#define rep(i,x,y) for(register int i=x;i<=y;i++)using namespace std;const int N=2e6+50;int n,m,s,mi,mx,ans,w[N],v[N],l[N],r[N];inline int read(){    int x=0,f=1;char ch=getchar();    while(!isdigit(ch)){
   if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}    return x*f;}inline int aabs(int x){
   return x>0?x:-x;}int sum[N],cnt[N];bool check(int mid){
   int y=0;    memset(sum,0,sizeof sum);    memset(cnt,0,sizeof cnt);        rep(i,1,n){        sum[i]=sum[i-1],cnt[i]=cnt[i-1];        if(w[i]>=mid)           sum[i]+=v[i],cnt[i]++;        }rep(i,1,m)        y+=(cnt[r[i]]-cnt[l[i]-1])*(sum[r[i]]-sum[l[i]-1]);    ans=min(ans,aabs(y-s));    if(y>s) return 1;    else return 0;}signed main(){    n=read(),m=read(),s=read();ans=mi=999999999999999999;    rep(i,1,n) w[i]=read(),v[i]=read(),mi=min(mi,w[i]),mx=max(mx,w[i]);    rep(i,1,m) l[i]=read(),r[i]=read();    int l=mi,r=mx;    while(l<=r)    {        int mid=(l+r)>>1;        bool FG=check(mid);                if(FG) l=mid+1;        else r=mid-1;    }    printf("%lld\n",ans);return 0;}
     

怀挺