poj3484 Showstopper 二分

　　题目地址

　　二分用的很是巧妙！关键是抽象出问题本质。

 #include <cstdio>
 #include <string>
 #include <cstring>
 const int maxn = ;
 #define ull unsigned long long
 ull X[maxn], Y[maxn], Z[maxn], C[maxn];
 ull N;
 ull judge(const ull &mid) {
     ull sum = ;
     for (int i = ; i < N; i++) {
         if (mid >= Y[i]) {
             sum += C[i];
         } else if (mid >= X[i]) {
             sum += ((mid-X[i])/Z[i]+);
         }
     }
     return sum;
 }
 char buf[];
 bool input() {
     int f = ;
     N = ;
     while ((f = ((gets(buf) != NULL))) && strlen(buf) > ) {
         sscanf(buf, "%llu %llu %llu", &X[N], &Y[N], &Z[N]);
         N++;
     }
     return f || N;
 }
 int main(void) {
     for (;input();) {
         if (!N) {
             continue;
         }
         ull last = ;
         for (int i = ; i < N; i++) {
             C[i] = (Y[i] - X[i])/Z[i] + ;
             last ^= C[i];
         }
         if (!(last & )) {
             printf("no corruption\n");
         } else {
             ull l = , u = 0xffffffff;
             for (;(u-l)>;) {
                 ull mid = (u+l) >> ;
                 if (judge(mid)&) {
                     u = mid;
                 } else {
                     l = mid +;
                 }
             }
             printf("%llu %llu\n", l, judge(l)-judge(l-));
         }
     }
     return ;
 }