; arr[i as usize][1] = y1 as i64; arr[i as usize][2] = y2 as i64; arr ][1] = y1 as i64; arr[(i + n) as usize][2] = y2 as i64; arr[(i + n) as usize][3 let mut dst = DynamicSegmentTree::new(max); let mut pre_x: i64 = 0; let mut ans: i64 = 0; for ::new(Node::new())), size: max, } } pub fn add(&mut self, ll: i64, rr: i64, cover , cur: Rc<RefCell<Node>>, l: i64, r: i64, ll: i64, rr: i64, cover: i64) { if ll <= l && rr >=
代码如下： use rand::Rng; fn main() { let len: i64 = 100; let value: i64 = 100; let test_time: ("测试结束"); } fn ways1(arr: &mut Vec<i64>, p: i64, x: i64) -> i64 { let mut sum = 0; for num in >, p: i64, x: i64) -> i64 { let mut sum = 0; for num in arr.iter() { sum += *num; , x: i64, num: i64, mod0: i64) -> i64 { // p/x 至少有几个 // (p % x) >= mod ? ) -> Vec<i64> { let mut ans: Vec<i64> = vec!
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("测试结束");}// 暴力方法fn sum1(n: i64) -> i64 { let mut cnt: Vec<i64> = vec! =n { ans += i * i * cnt[i as usize]; } return ans;}fn get_sqrt(n: i64) -> i64 { let mut l: i64 = 1; let mut r = n; let mut m: i64; let mut mm: i64; let mut ans = 1; while l , n: i64) -> i64 { let r = cover(v, n); let l = cover(v, n + 1); return ((r * (r + 1) * ((r << 1) + 1) - l * (l + 1) * ((l << 1) + 1)) * n) / 6;}fn cover(v: i64, n: i64) -> i64 { let mut l =
[ (Datum::I64(1), Datum::I64(2), Datum::U64(2)), ( Datum::I64(i64 ::MIN), Datum::I64(1), Datum::U64(i64::MIN as u64), ), ( Datum::I64(i64::MAX), Datum::I64(1), Datum::U64(i64::MAX [ (Datum::I64(-1), Datum::I64(2)), (Datum::I64(i64::MAX), Datum::I64(i64::MAX)) , (Datum::I64(i64::MIN), Datum::I64(i64::MIN)), ]; for (left, right) in cases
8 ; 调用 @selector(length) 方法 %5 = tail call i64 bitcast (i8* (i8*, i8*, ...)* @objc_msgSend to i64 i64 (i8*, i8*)*)(i8* %2, i8* %8) #3, ! 8 ; 准备参数 NSRange ,分别是 {length-1,1} %12 = insertvalue [2 x i64] undef, i64 %10, 0 %13 = insertvalue [2 x i64] %12, i64 1, 1 ; 获取子串 %14 = tail call %1* bitcast (i8* (i8*, i8*, ...)* @objc_msgSend to %1* (i8*, i8*, [2 x i64])*)(i8* %2, i8* %11, [2 x i64] %13) #3, !
("ans = {}", ans); } const mod0: i64 = 1000000007; fn total_strength(arr: &mut Vec<i32>) -> i32 { let n = arr.len() as i32; let mut pre_sum = arr[0] as i64; let mut sum_sum: Vec<i64> = vec! []; for _ in 0..n { sum_sum.push(0); } sum_sum[0] = arr[0] as i64; for i in 1 []; for _ in 0..n { stack.push(0); } let mut size: i32 = 0; let mut ans: i64 >, l: i32, m: i32, r: i32) -> i64 { let left = (m as i64 - l as i64) * (sum_sum[(r - 1) as
("ans = {}", ans); } const mod0: i64 = 1000000007; fn total_strength(arr: &mut Vec<i32>) -> i32 { let n = arr.len() as i32; let mut pre_sum = arr[0] as i64; let mut sum_sum: Vec<i64> = vec []; for _ in 0..n { sum_sum.push(0); } sum_sum[0] = arr[0] as i64; for i in 1 []; for _ in 0..n { stack.push(0); } let mut size: i32 = 0; let mut ans: i64 >, l: i32, m: i32, r: i32) -> i64 { let left = (m as i64 - l as i64) * (sum_sum[(r - 1) as
代码如下： use rand::Rng; fn main() { let nn: i64 = 18; let kk: i64 = 20; let test_time: i32 = ("测试结束"); } // 暴力方法 // 为了验证 fn mod_ways1(s: &str, k: i64) -> i64 { let n = s.len() as i64; let } } return ans; } // 正式方法 // 时间复杂度O(N * k) fn mod_ways2(s: &str, k: i64) -> i64 { let mut cur: Vec<i64> = vec! []; for _ in 0..k { cur.push(0); } // 帮忙迁移 let mut next: Vec<i64> = vec!
llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %222 = llvm.mlir.constant(0 : index) : i64 %223 = llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %236 = llvm.bitcast %235 : !llvm<"double*"> to ! llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %238 = llvm.bitcast %237 : !llvm<"double*"> to ! llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %240 = llvm.bitcast %239 : !llvm<"double*"> to ! , [2 x i64], [2 x i64] } %8, 0 %104 = mul i64 %96, 2 %105 = add i64 0, %104 %106 = mul i64 %100
("测试结束"); } // 时间复杂度O(N * V)的方法 // 为了验证 fn max_sum1(arr: &mut Vec<i32>) -> i64 { let n = arr.len in arr.iter() { max = get_max(max, *num); } let mut ans = 0; let mut dp: Vec<Vec<i64 >>) -> i64 { if p <= 0 || i == -1 { return 0; } if dp[i as usize][p as usize] ! + next; dp[i as usize][p as usize] = ans; return cur as i64 + next; } // 正式方法 // 时间复杂度O(N) { n = get_min(max, n); ((max as i64 * 2 - n as i64 + 1) * n as i64) / 2 } // 为了验证 fn random_array
下面给出代码： i64 POW(i64 a,i64 b,i64 mod) { i64 ans=1; while(b) { if(b&1) ans=ans*a%mod exGcd(i64 a,i64 b,i64 &x,i64 &y) { i64 t,d; if(! reverse(i64 a,i64 b) { i64 x,y; exGcd(a,b,x,y); return (x%b+b)%b; } i64 C(i64 n,i64 m,i64 (i64 n,i64 p,i64 t) { if(! C2(i64 n,i64 m,i64 p,i64 t) { i64 x=POW(p,t); i64 a,b,c,ap=0,bp=0,cp=0,temp; for(temp=n;
代码如下： use rand::Rng; fn main() { let nn: i64 = 20; let kk: i64 = 30; let test_time: i32 = ("测试结束"); } // 动态规划 fn ways1(nums: &mut Vec<i64>, k: i64) -> i64 { let n = nums.len() as i64; >, k: i64) -> i64 { let mut res = 1; let mut i: i64 = 0; let mut j: i64 = 0; while i , b: i64) -> i64 { if a == b { return 1; } let mut x = 1; let mut y = 1; (n: i64, k: i64) -> Vec<i64> { let mut ans: Vec<i64> = vec!
i32)))\n (type $1 (func (param i32 i64 i64 i32 i32)))\n (type $2 (func (param i32 i64 i32 i32)))\n (func (param i64)))\n (type $7 (func (param i32 i32)))\n (type $8 (func (param i64 i64 i64 i64) (result i32)))\n (type $9 (func (param i64 i64 i64 i64 i32 i32) (result i32)))\n (type $10 (func (param i32 i32 i64)))\n (type $18 (func (param i64 i64 i32 i32)))\n (type $19 (func (param i32 i64 i32) (result (func $27 (param i64 i64 i64 i64) (result i32)))\n ......
, 0, (MM_PAGE_PRIORITY) 0x40000010); v26 = * v18; v27 = 0 i64; v28 = 48; v29 = 0 i64; v31 = 0; v30 = 0 i64; _mm_storeu_si128((__m128i * ) & v32, (__m128i) 0 i64); v4 = (unsigned int) ZwOpenProcess %x\n", v34, v4); v21 = 0 i64; v33 = 0 i64; v35 = v18[4]; if ((int) v4 >= 0) { v4 = (unsigned [1] = 0 i64; } 在我深入研究欺骗者之前，我只想说明 amlegit 的开发人员没有发现一种欺骗 hwid 的新方法。 \n"); spoof_nic(0 i64); debug_with_prefix((__int64) "SMBIOS...
("ans = {}", ans);}fn people_aware_of_secret(n: i32, delay: i32, forget: i32) -> i32 { let mod0: i64 = 1000000007; // dpKnow[i], 第i天知道秘密的人 let mut dp_know: Vec<i64> = repeat(0 as i64).take((n + 1 ) as usize).collect(); // dpForget[i], 第i天将要忘记秘密的人 let mut dp_forget: Vec<i64> = repeat(0 as i64 ).take((n + 1) as usize).collect(); // dpShare[i], 第i天可以分享秘密的人 let mut dp_share: Vec<i64> = repeat (0 as i64).take((n + 1) as usize).collect(); // 第1天的时候，知道秘密的人1个，A // 第1天的时候，将要忘记秘密的人0个 // 第1
("ans = {}", ans);}fn selling_wood3(m: i64, n: i64, prices: &mut Vec<Vec<i64>>) -> i64 { // dp表！ let mut dp: Vec<Vec<i64>> = vec![]; for i in 0..m + 1 { dp.push(vec!
llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %222 = llvm.mlir.constant(0 : index) : i64 %223 = i64 %226 = llvm.mlir.constant(1 : index) : i64 %227 = llvm.mul %219, %226 : i64 %228 = llvm.add llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %236 = llvm.bitcast %235 : !llvm<"double*"> to ! llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %238 = llvm.bitcast %237 : !llvm<"double*"> to ! llvm<"{ double*, i64, [2 x i64], [2 x i64] }"> %240 = llvm.bitcast %239 : !llvm<"double*"> to !
int i, err := strconv.Atoi(str) if err == nil { fmt.Printf(“i: %v\n”,i) } // string 转 int64 i64 ,err := strconv.ParseInt(str,10,64) if err == nil { fmt.Printf(“i64: %v\n”,i64) } // string 转
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