看到绝对值最小这种逼近的问题，应该能想到二分。
容易发现，随着W的变大，符合条件的矿石越来越少，即满足单调关系。因此我们二分W，然后在W确定时求出x = ΣYi，如果x >= S，说明W取小了，向右二分；否则向左二分。每一次都更新答案。
求x用前缀和就解决了。
时间复杂度：二分O(logn)，求值O(n)，总共O(nlogn)。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 2e6 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m; 38 ll S; 39 struct Node 40 { 41 ll w, v; 42 }t[maxn]; 43 struct Node2 44 { 45 int L, R; 46 }q[maxn]; 47 48 int num[maxn]; 49 ll sum[maxn]; 50 ll calc(ll x) 51 { 52 for(int i = 1; i <= n; ++i) 53 { 54 if(t[i].w >= x) 55 { 56 num[i] = num[i - 1] + 1; 57 sum[i] = sum[i - 1] + t[i].v; 58 } 59 else 60 { 61 num[i] = num[i - 1]; 62 sum[i] = sum[i - 1]; 63 } 64 } 65 ll ret = 0; 66 for(int i = 1; i <= m; ++i) 67 ret += (num[q[i].R] - num[q[i].L - 1]) * (sum[q[i].R] - sum[q[i].L - 1]); 68 return ret; 69 } 70 71 ll ans = (ll)INF * (ll)INF; 72 73 int main() 74 { 75 n = read(); m = read(); S = read(); 76 ll l = INF, r = -INF; 77 for(int i = 1; i <= n; ++i) 78 { 79 t[i].w = read(), t[i].v = read(); 80 l = min(l, t[i].w); 81 r = max(r, t[i].w); 82 } 83 for(int i = 1; i <= m; ++i) q[i].L = read(), q[i].R = read(); 84 while(l < r) 85 { 86 ll mid = (l + r + 1) >> 1; 87 ll x = calc(mid); 88 if(x <= S) r = mid - 1, ans = min(ans, S - x); 89 else l = mid, ans = min(ans, x - S); 90 } 91 write(ans), enter; 92 return 0; 93 }