Codeforces 397A On Segment's Own Points(遍历)
2014-02-27 13:34
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题目链接:Codeforces 397A On Segment's Own Points
题目大意:给出n个区间,第一个区间为自己的区间,然后n-1个区间为别人的区间,问说你有多少长度的区间是自己独享的。
解题思路:水题,因为r的上限为100,所以完全可以开个100的数据记录每段区间是否被占用。
#include <stdio.h>
#include <string.h>
const int N = 105;
int n, v
, L, R;
int main () {
memset(v, 0, sizeof(v));
scanf("%d%d%d", &n, &L, &R);
int x, y;
for (int i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
for (int j = x + 1; j <= y; j++) v[j] = 1;
}
int ans = 0;
for (int i = L + 1; i <= R; i++) if (!v[i])
ans++;
printf("%d\n", ans);
return 0;
}
题目大意:给出n个区间,第一个区间为自己的区间,然后n-1个区间为别人的区间,问说你有多少长度的区间是自己独享的。
解题思路:水题,因为r的上限为100,所以完全可以开个100的数据记录每段区间是否被占用。
#include <stdio.h>
#include <string.h>
const int N = 105;
int n, v
, L, R;
int main () {
memset(v, 0, sizeof(v));
scanf("%d%d%d", &n, &L, &R);
int x, y;
for (int i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
for (int j = x + 1; j <= y; j++) v[j] = 1;
}
int ans = 0;
for (int i = L + 1; i <= R; i++) if (!v[i])
ans++;
printf("%d\n", ans);
return 0;
}
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