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 = 100000;
#define ull unsigned long long
ull X[maxn], Y[maxn], Z[maxn], C[maxn];
ull N;
ull judge(const ull &mid) {
    ull sum = 0;
    for (int i = 0; i < N; i++) {
        if (mid >= Y[i]) {
            sum += C[i];
        } else if (mid >= X[i]) {
            sum += ((mid-X[i])/Z[i]+1);
        }
    }
    return sum;
}
char buf[1024];
bool input() {
    int f = 0;
    N = 0;
    while ((f = ((gets(buf) != NULL))) && strlen(buf) > 2) {
        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 = 0;
        for (int i = 0; i < N; i++) {
            C[i] = (Y[i] - X[i])/Z[i] + 1;
            last ^= C[i];
        }
        if (!(last & 1)) {
            printf("no corruption
");
        } else {
            ull l = 0, u = 0xffffffff;
            for (;(u-l)>0;) {
                ull mid = (u+l) >> 1;
                if (judge(mid)&1) {
                    u = mid;
                } else {
                    l = mid +1;
                }
            }
            printf("%llu %llu
", l, judge(l)-judge(l-1));
        }
    }
    return 0;
}