noip 2012 提高组 day2 部分题解

　　这道题有多种解法，我用的是扩展欧几里得算法求到的答案

 #include<iostream>
 #include<fstream>
 #include<cstdio>
 using namespace std;
 typedef long long ll;
 ifstream fin("mod.in");
 FILE *fout = fopen("mod.out","w");
 void gcd(ll a,ll b,ll& x,ll& y){
     if(!b){    x = , y = ;    }
     else{    gcd(b, a%b, y, x);    y -= x * (a / b);    }
 }
 ll a1,b1;
 ll x1 = ,y1 = ;
 int main(){
     fin>>a1>>b1;
     gcd(a1, b1, x1, y1);
     fprintf(fout,"%ld",(x1 + b1 * )%b1);
     return ;
 }

---

　　这道题把第i个人看做一个有序的序列（1、2、3、4....）然后二分

至于求和，就像这么处理：

　　、

接着从前面开始求和。。。

就像这样可以求出每一天的教室使用量，如果1 ~ v天中有哪一天不够用了，就在前半段

查找，如果都足够，就向后面查找，每次不够的时候更新结果result

 #include<iostream>
 #include<fstream>
 #include<cstdio>
 #include<cctype>
 #include<cstring>
 using namespace std;
 typedef bool boolean;
 FILE *fout = fopen("classroom.out","w");
 template <class T>
 inline void get(T &u){
     char x;
     while(!isdigit(x=getchar()));
     for( u=x-; isdigit(x=getchar()); u*=,u+=(x-));
     ungetc(x,stdin);
 }
 int *d;
 int *s;
 int *t;
 int *r;
 int *buf;
 int m,n;
 boolean solve(int v){
     memset(buf, , sizeof(int) * (n + ));
     int sum = ;
     int limit = ;
     for(int i = ;i <= v;i++){
         buf[s[i]] += d[i];
         buf[t[i] + ] -= d[i];
         limit = max(limit, t[i]);
     }
     for(int i = ;i <= limit;i++){
         sum += buf[i];
         if(sum > r[i])    return true;
     }
     return false;
 }
 int main(){
     freopen("classroom.in","r",stdin);
     get(n);
     get(m);
     r = new int[(const int)(n + )];
     d = new int[(const int)(m + )];
     s = new int[(const int)(m + )];
     t = new int[(const int)(m + )];
     buf = new int[(const int)(n + )];
     for(int i = ;i <= n;i++)    get(r[i]);
     for(int i = ;i <= m;i++){
         get(d[i]);
         get(s[i]);
         get(t[i]);
     }
     int from = ;
     int end = m;
     int result = ;
     while(from <= end){
         int mid = (from + end) >> ;
         if(solve(mid)){
             result = mid;
             end = mid - ;
         }else from = mid + ;
     }
     if(!result)    fprintf(fout,"0\n");
     else fprintf(fout,"-1\n%d\n",result);
     return ;
 }